I am confused on tonight's homework from 5.2 on the word problems. I get the SETUP of the formula and what you plug in for the first value, but not for the e. If someone can please explain the concept of "e and ln", that would be great. I completely understand logs, but I seem to have trouble as to figuring out when to use the e and ln.
Ln and e both apply to natural logs. When using e in a formula you do not need to plug anything in, e is a constant. On ur calculator it is with the button for division. So just plug the equation into your calculator
Hey Chandan you do not need to plug anything in for e, like Alex has said. Then once you find out what variable to plug in you can just enter the entire equation into your calculator. If you ever need help just go into the ARC durring lunch. It is ALOT of help. I go almost everyday, and I get most if not all of my homework done.
yeah Chandan! I used the calculator button and it really helped me. 2nd LN gets you to e^x, and you can just enter it in from there. Hope it helps!
For section 5.2 #34, the equation is f(x)=!+3^-(x-4). Shouldn't this simplify to 1+3^(-x+4)? When I type both of these into my calculator, they are equal but x is moved 4 to the right. Can anyone explain why it isn't 4 to the left?
I was wondering if somebody could help me with problem 58 in section 5.2?
I'm a little confused on #56 from 5.2. I thought it you had the same base you could set the exponents equal to each other but it doesn't seem to work in this problem. Thanks!
Kayla - Hopefully it helps. Since you are adding the 4 to the negative value of x you are causing whatever the x value is to act as if it is 4 bigger then its negative ( 1 -> -1 +4 = 3 or 5 -> -5 +4 =-1) Since you are adding a positive four, the x is affected (effected?) in a positive way thus moving to the right (positive) on the graph. That might make no sense, but yes. Nithya - 58 is a log problem. So if you look ahead to 5.3 you will find the theorem on how to solve with logs and get (log3 / log2) = x then just plug in the x to 4^-x
Lindsey - The part that might be throwing you off in 56 is the (e^4)^x . This is the same as e^4x. Once you get that all the bases are the same and you get the equation 4x + x^2 = 12 and with some movement it makes x^2 +4x -12 which facters into (x+6)(x-2)giving you the answers
Nithya, if you don't want to use logs, just break 4^-x down to 2^2(-x) which simplifies to 2^-2x. You know 2^x=3. 2^-2x would be 3^-2 (because to get 2^-2x from 2^x, you multiply the exponant by -2). Then, multiply 3's exponant (1) by -2, getting 3^-2. This gives you 1/(3^2) which simplifies to 1/9.
Ok Jamez, I now understand 34. However, what's up with 42's graph? Why is it a straight line and is the domain/range all reals without any HASYs?
I did not get just a straight line, I got a curved graph that went up and to the right with a hasy of 7. The domain is all reals but the range is y<7
Yeah, for number 42, since it's 7-3e^(-2X) it's just moved up 7 but their is a still a curve. Can someone explain to me, how to do number 52. I moved the 2^X to the right and got 4^X=2^X, but I'm stuck.
4^x is equal to 2^2x. Then you have the same base and can set 2x = x
I am just clarifing, is a hole the parrt you cancel off when simplifing the formula. The cancelled part is set to equal zero and then solve and that will be your hole.
Hey Angelica for #52 what I did was move over the 4^x and then you can set the two to exponents equal to eachother so you end up having x=2x then when your solving it you would move the x over and set it equal to zero and you just end up having x=0. hope that helped!Can someone explain how to di #60? I tried using the method that Jamez said by using the logs but I think I'm entering it wrong in my calculator..or just not usin the right method
I'm having trouble with number 60 in 5.2 how would you set up the problem? And do you do it with or without logs?
For #60 i made 5^-x 1/5^x then i multiplied 5^x by each side and then the equation was 1 = (3)(5^x) i then divided by 3 and had 1/3 = 5^x From there i cubed each side and got 1/27 = 5^3x
Here is what I did for question #60 in section 5.2.5^-x = 3x = log(3)/-log(5)5^3x = ?5^3(log(3)/-log(5)) = 1/27
Could someone help me with setting up number 48 on the 5.2 homework please? I'm probably over-thinking it, it seems rather simple just looking at it.
Actually I havit. Never mind :)
To solve 60 without logs, I did:5^(-x) = 35^x = 1/35^x^3 = (1/3)^35^(3x) = 1/27
Thanks guys, umm 50 and 56 anyone?
Just kidding about 56
Number 50:(1/2)^(1-x) = 4(1/2)^(1-x) = 2^2(1/2)^(1-x) = (1/2)^-21-x = -23 = x
Does anyone know how to get started on problems 42 and 45 in chapter 5.3?
Can someone hlep me with #37 on 5.3, would it just be x isgreater than 3 or can the part in the parenthesis be negative?
Jose - for number 54 you must remember that a the argument of a log must be greater than 0. Also, the function has a verticle asymtote at x=0 because it makes the problem undefined. Thereby making there 3 possible intervals for numbers to fall under:Negative infinity to -1-1 to 00 to infinity42 is not a part of the homework, however the same rule of the argument for the log being grater than 0 still applies. So you can set up the question as:x^2-1>0This can be factored into (x+1)(x-1)>0then test the intervals of:Negative infinity to -1-1 to 11 to infinityAlex-Ln is just another form of log. therefore the rule about the arguement being greater than 0 still applies. so set x-3>0 and solve for x.
How do you do problems like number 33?
Thats 45 not 54...
Use the general form for logs such that:Y=log(whatever base it is)x, changing to:x=(base) raised to the Y power. so for 33 you can convert it to:(Rt 2)^x = 4(Rt 2)^x = (Rt 2)^4x=4
for the ones like 33 you set it equal to x because then you can turn it into a exponent type problem then isolate x and sovle for it.
For number 41 from 5.2, I tried to solve it and I got a weird domain and I'm not really sure what to do so can someone help?
41) For the sake of arguement, lets say that at any time i say log for this problem it is assumed to be base 1/2I said log(x^2-2x+1) can be factored into:log(x-1)^2Which is equal to:log(x-1) + log(x-1)Since each log needs to have a positive value and they're equal, x must be greater than 1
Could anyone show me how to do problems 94 and 97 in section 5.3? especially part b.
thanks jake, and melina for 94b you have to use logs so that the inverse would be log5x = y (where the 5 is the base) because you have to switch x and y to find the inverse. For 97b you first switch x and y, then subtract 1 to get x-1=2^(y-3). Then you take a log so you end up with log2(x-1)=y-3. Then you just add 3 to get your final answer of log2(x-1)+3=y (where 2 is the base). I hope that helps
Jake and Justin, but for problem 41, the answer in the book says x does not equal 1, not x>1. Why is this?
x does not equal 1 is correct. Since you are taking the log (x-1)^2, you can plug in a number less than 1 for x and still get a positive number. For example, if you plug in -1 for x, then you get (-1-1)^2, which equals (-2)^2 which equals 4. Since that is positive, all number less than 1 are in the domain as well as the numbers greater than one. The only exclusion in the domain is when x is 1, because you cannot take the log of 0. I hope that makes sense.
Kayla, i think it's because theres a vasy there!
Just kidding, Kelly is right!
Can anyone show me how to do number 105 in section 5.3?? I don't understand how you can solve for h.
Numbers 80 and 84 in 5.3?
Number 100 on 5.3?
hey lexy! this is what i did. Since D=miligrams and the problem states what is the time in between injections when the miligrams is 2 I put 2 in for D n the equation. The equation then looks like 2=5e^-0.4h. From this equation convert it to a natural logs problem and it then looks like ln(2/5)=-0.4h and then you can figure out what h is and it equals around 2.29. Since the problem is asking for the time in hours the answer would be 2 hours and around 17 minutes.
Alex, ok for 100 a) I just plugged in the number they gave you (.0000001) for H. Then your equation is -log(.0000001) which equals 7. So your pH is 7.
hey lindsay i know how to do #84. You can change the original problem to 6^5x+3=36. Now you need to get an equal base in order to solve so you turn 36 into 6^2. Now that you have equal bases it looks like 6^5x+3=6^2. You then make the powers equal to eachother which looks like 5x+3=2 and from there you solve for x and it equals 1/5.
Can someone please help me on problem 105 in 5.3? I'm really confused on how to solve it.
can someone show me how to do #45 and #108 from 5.3
For #45 in 5.3 we're finding the domain of log(subscript 5)((x+1)/x)So we first know that x ≠ 0 since we can't divide by 0.Then if you set x + 1 = 0 x can't equal -1 So Domain: 0<x<-1
Can someone please do number 78in section 5.3 please
For 78, you start with the original equation of log (base x) 1/8 = 3. You rewrite this in exponential from, so you have x^3=1/8. Next, you rewrite the 1/8 as (1/2)^3. If you substitute this in, you have x^3= (1/2)^3. Take the third root of each side and you have x=1/2. Hope that helps!
For 100 b, you take the given pH, 4.2, and simply plug it into the given equation. It should look like 4.2 = -log H. Multiply each side by negative one to isolate the log: -4.2=log H. Rewrite this in exponential form, so you should have 10^(-4.2) = H. Then, simply plug this into your calculator and you get H = .0000631. I’m doing this without my book and just using my spiral so sorry if there are any mistakes in the notations or anything.
Could someone help me with number 5 and 9 please? I don't know where to start with number 5, and i got as far as ln8^(1/3) on number 9.
For 5, you take ln (2e) and separate it into ln 2 + ln e. It was given that ln 2 = a, so you have a + ln e. ln can be rewritten as log (base e). So if you take the log (base e) e, the bases cancel and that equals 1. So for 5, the answer is a + 1. For 9, you can rewrite the ln (5th root of 18) as ln 18^(1/5). You can move the exponent to the front, so you have 1/5 ln 18. Next you factor 18, so you get 1/5 ln (3*3*2). Using that addition property, you get 1/5 (ln 3 + ln 3 + ln 2). You can substitute the variable a and b since it is given, so you get 1/5 ( b + b + a). Keep combining: 1/5 (2b + a). Distribute the 1/5 and you get 2/5a + 1/5b. Let me know if these are right. I didn’t check them in the book.
Sorry, i meant you get 2/5b + 1/5a, not 2/5a and 1/5b.
Oh okay, thanks Kelly! I get it now
Lindsey: for 80 on 5.3, since you have lne^-2x=8, you would first disregard the e, as the ln of e cancels out. Then you're left with -2x=8. By dividing both sides by -2, your answer would equal, x=-4. Anika: The equation given is D=5e^(/.4h). As we have to find what h (the hours after the drug is administered) equals when D=2, we substitute 2 for D. This gives us:2=5e^(-.4h). Then divide both sides by 5. This results in:2/5=e^(-.4h). By finding the natural log (ln) of both sides, the new equation is:ln(2/5)=-.4has the natural log and e cancel out. By dividing both sides by -.4, you find h to equal 2.29 hours or 2 hours and 17 minutes. After 2 hours and 17 minutes, the drug has to be re-administered. I hope that helps:)
for 5.4 problems 43 & 44 how do you plug that into the calculator?
#43using change of base you getlog(16)/log(e^(2)=about 1.39then you have to go into the catalog (which is second 0) and find eand then you just do e^1.39 and you get about 4#44 is the same waylog(9)/log(e^(2))=about 1.1e^1.1= about 3i hope that helps
Can someone please show me how to do number 25 in section 5.5?
Could some explain to me how to do number 13 and 21 in section 5.5???
So for 25, you take the log of each side, so you get log 1.2^x = log .5^-x. Bring the exponents to the front : x log 1.2 = -x log .5. Subtract (–x log .5) from each side, to get: x log 1.2 – (-x log .5) = 0. Factor out the x from each log: x(log 1.2 – (-log .5)) = 0. Divide each side by (log 1.2 – (-log .5)): x=0.
For 13, start by rewriting the first term as (2^x)^2. When you substitute this in for the first term, your new equation looks like this: (2^x)^2 + (2^x) – 12 = 0. From here, you treat it like a quadratic equation. Factor it to get: ((2^x) + 4)((2^x) – 3) = 0. Next, you set each factor equal to 0. For the first factor, you get 2^x = -4. You can eliminate this as a possible solution because 2^x cannot be a negative number. When you set the second factor equal to 0, you get 2^x = 3. Rewrite this in log form to get x= log (base 2) 3. Use the change of base formula to plug this into the calculator: log 3/log 2 = 1.585.
And for 21, start by taking the log of each side: log 3 ^(1-2x) = log 4^x. Bring the exponents to the front: (1-2x) log 3 = x log 4. Multiply each side by (1/x) to eliminate the x on the right side of the equation: (1/x)(1-2x) log 3 = log 4. Divide each side by log 3 to isolate the terms with x in them: (1/x)(1-2x) = log 4/log 3. Distribute the (1/x) on the left side of the equation: (1/x) – (2x/x) = log 4/log 3. The x’s cancel in the (2x/x) term, so you get: (1/x) – 2 = log 4/log 3. Add 2 to each side: 1/x = log 4/log 3 + 2. Multiply each side by x: 1 = x(log 4/log 3 + 2). Divide each side by (log 4/log 3 + 2) to isolate x: 1/(log 4/log 3 + 2) = x. When you plug that into a calculator, you get x = .307. This is kind of confusing so if you have any questions about my work just let me know!
For 29 in 5.5, could you divide by 5 first to make it <2^3x=(8/5)>? I'm not sure though, because then you don't have 2 and 8 having the same base. But I guess you could do the ln thing they did in example 6.
Muhammed, you can divide by 5 because it is outside the parentheses, then you take the log of each side, move the exponent for the 2^3x side and do rewrite log(8/5) as log8-log5. Then solve for x. Can somebody explain #71 on 5.4? I don't understand how they get 1 because I keep ending up with logs in my answer...
can somebody explain 14 and 22 on 5.5?
To solve number 14, first:change 3^2x to 3^(x^2). That way, you can understand that when they wrote 3^2x, they also meant that it is the same thing as 9^x (basically, they sort of already simplified it for you). It should have originally looked like:9^x + 3^x - 2 but the current equation should now look like:3^(x^2) + 3^x - 2 Which now looks like a quadratic equation so you factor to get: (3^x + 2)(3^x - 1) If you're confused on how i got this, look back to this: 9^x + 3^x - 2Next, set each equal to zero and you should know that you cant deal with the equation on the left because it has no solution. For the equation on the right, it should look like:3^x = 1 (after you set it to zero)By one of the log properties, you should know that this equation is the same as:x = log base 3 of 1To solve, you use the change of base property and end up getting log of 1 divided by log of 3.The answer is X = 0.
im kind of confused on number 26? could someone please explain it. thank you
I don't get at all how to do numbers 21 and 25 from 5.5. Could someone explain it?
Could someone please show me how to do numbers 10, 34, 38, and 42 from section 5.5?
For number 21, you have to use the natural log, so with the 3^1-2x on the left side, you bring the 1-2x out front so it's (1-2x)ln3, and then do the same for the 4^x, so then the equation looks like (1-2x)ln3 = (x)ln4. Then you distribute, so it's now ln3-2xln3=xln4. Then you have to get the x's on one side, so it's ln3=2xln3-xln4. Because you're trying to solve for x, you have to factor, so then it becomes ln3=x(ln4+2ln3). Then you divide and the result should be ln3/ln4+2ln3. I hope that makes sense :)
Yea that helped Sejzelle. Thanks.
For number 25, you also use the natural log. so 1.2^x=(0.5)^-x becomes xln1.2=-xln0.5. Again, because you're solving for x, bring the x terms to one side, factor, and then divide. So it becomes xln1.2+xln0.5=0. But because there's a zero on the end, whatever we divide it by, it's always going to be zero, so you can just stop there...the answers is zero.
Sohaila, for number 26, you have to use the natural log, so the equation becomes (1+x)ln(0.3)=(2x-1)ln(1.7). Then you have to distribute on both sides, so it becomes ln(0.3)+xln(0.3)=(2x)ln(1.7)-(x)ln(0.3). Then you have to get the x terms on the same side, so by subtracting the xln0.3 and adding the ln1.7, the whole thing becomes ln1.7+ln0.3=2xln1.7-xln.3. Since we're trying to get x by itself, you have to factor it so you can then divide, so then it's x(2ln1.7-ln0.3). Divide the ln1.7+ln0.3 by what is in the parentheses and you get x=-.297
Hey, does anyone know how to solve this:4^(2x)-4^(x+1)-5
Sorry, it's 4^(2x)-4^(x+1)-5=0
For the extra credit problem, the answer is x=(log5)/(log4)-1. By using the calculator, the final value of x is .16 or 4/25. The original equation is 4^ (2x)-4^ (x+1)-5=0. Then I combined like terms, resulting in 4^(x+1)-5=0.Then I added 5 to both sides: 4^(x+1)=5.Then I found the log (with a base of 4) for both sides:log4^x+1)/log4=log5/log4simplified: x+1=log5/lo4Then I sutracted both sides by 1:x=(log5/log4)-1.
to the extra credit problem its 1.16 if you use the calculator or log5/log4 without a calculator
Wait sorry! my explanation failed:Pso the answer is x=(log5)/(log4)And here's how it's actually found:The original equation is 4^ (2x)-4^ (x+1)-5=0. Then I combined like terms, resulting in 4^(x+1)-5=0.Then I added 5 to both sides: 4^(x+1)=5.Then I found the log (with a base of 5) to both sides:log4^(x+1)/log5=log5/log5 Then I brought the exponent down:(x+1)log4/log5=1 Then I divided both sides by (x+1):log4/log5=1/(x+1)Then, in order to get rid of the fraction, I raised both sides to the -1. (x+1)=(log5/log4)^-1Then I brought the exponent down:(x+1)=-1(log5/log4Then I subtracted 1 from both sides:x=-(log5/log4)-1Since, with logs, subtractions is the same as division, I divided:x=-1(log5/log4)/-1After simplifying:x=log5/log4
EXTRA CREDIT IS 1.16!!!!!!!
By the way, for number 31 on the homework, what do you do after you have log base a of (((X^2)+2x-3)/((X^2)+4x-12))
Oh I got it. Thanks Kayla
And yea I got (log 5/log 4) for the EC too
I know that this problem was never assigned, but could someone explain how to do #7 from 5.5?
Here's number 7 Kyle...First you need to bring the 3 up as an exponent in the first term. Then you can change the two terms into one by multiplication. So it would be log base 2(x-1)^3 *4=5. Then you can change it to an exponent so 2^5= (x-1)^3*4. Then you can divide by 4 on both sides and your left with 8= (x-1)^3. You raise both sides to the 1/3 to get rid of the exponent and you get that x=3! Hope you can understand all that!
Hey can someone help me solve number 22 from section 5.5? Its: 2^(x+1)=5^(1-2x)
Whoo got the e.c. 1.16! But on the homework in 5.5, number 35 I'm not sure if I found the common base correctly but I ended up with log(base)2 (x+1)squared/x=2. Is that right?
I had a quick question on #47 in 5.5i figured it out with my dad but I'm still not understanding why. The problem is "log(base)5(x+1)-log(base4)((x-2)=1"Could someone please work out this problem? I'm confused at the steps to solving this...
Beatrice-For number 47 it is a problem that you plug into your graphing calculator.
Hey, I know some people already asked this but as far as I can tell it hasn't been answered yet; I'm a little confused with #22 on the 5.5 homework, can anyone help?
Alec-For number 22 start by taking the log of both sides, so you getlog 2^(x+1) = log 5^(1-2x)(x+1)log 2 = (1-2x)log 5distribute the logsxlog2 + 1log2 = 1log5 - 2xlog5move so that x's are on one sidexlog2 + 2xlog5 = 1log5 - 1log2take out the x'sx(log2 + 2log5) = log5 - log2divide by (log2 + 2log5)x = (log 5 - log 2) (log 2 + 2log5)
Just a quick questoin: If there's a negative sign before a log such as -log12 or anything like that, it'd be the same as log12^-1, right? So the negative sign is just a -1, right?
Ratuja, you're right.-log12 = log12^-1
Just to add on to what I said before:-1log12, take the -1 and make it the exponent, so you would have log12^-1:]
For question 62 in the review. After you take the log of both sides in the equation do you take the x^2 and x^3 to the front of the log or just the 2 and 3?
Cody,Yes, you take the X^2 and X^3 to the front of the logs leaving you with:X^3ln2 = X^2ln3
I am having trouble finding the answer to number 38 in 5.5 can anyone help. The problem is: log(x)/log(9)+ 3 log(x)/log(3)= 14
For number 38, x=81.The original equation is: logx/log9+3logx/log3=14.Since the bases aren't the same, you need to apply change of base. To change the base to three, you do the following for logx/log9.logx/log9= (logx/log3)/(log9/log3).Then you simplify the denominator:log9/log3=log3^2/log3. The 3s cancel, and you're left with two for the denominator. You're equation now is:(logx/log3)/2 + 3(logx/log3)=14To get rid of the fraction, multiply through by 2:logx/log3 + 6(logx/log3)=28. To combine the terms, 6 becomes the exponent to the second term:logx/log3 + (logx/log3)^6=28Multiply the left side of the equation:logx^7/log3=28bring the 7 down as a coefficient:7logx/log3=28divide both sides by 7:logx/log3=4transform to exponential form3^4=xx=81Hope that helps:)
So can someone please explain tome #65 in the review!! Its a wordproblem, the one that is titled amplifying sound. its says thatthe formula is P=25e^.1d but i don't know how to find d...
Julianne-first of all, you set p equal to 50 and then divide both sides to get2=e^.1dthen take the natural log of both sides to get:ln2=lne^.1dthe lne cancels out, so you're left with ln2=.1dthen you just divide ln2 by .1 with your calculator to find d hope that helps
Can someone please help me with number 55 on tonights review hw. I don't get how it..
Can anyone explain how to do that inverse problem from the review problems we did in class?
While i was reviewing for the test, i got stuck on #40 on pg 328, can anyone help?
Wesley-For number forty, i looked at it that log(base 2)of x is x so it would be x to the x is 4 which i know is 2. Then i said 2 equals log(base 2)of x and then exponential equation it and you get 2 to the 2 equals x. So x equals 4
Kyle- here is what I did for number 55:8 = 4^(x^2) x 2^(5x)First you change all the bases of the exponents to make them versions of 2. 8 becomes 2 to the third power, 4 becomes two squared, and 2 is still just two. So now the problem is : 2^3= (2^2x2) x (2^5x). Since all the bases are the same, the big 2 falls out: 3 = 2x^2 + 5x : and the 2xsquared and the 5x can be added because the had the same base (it’s a property of exponents). So now you move everything to the same side to get: 0= 2xsquared + 5x + 3. Solve using the quadratic formula (or factoring) to get x= 0.5 and -3.
Mihir- are you talking about the one that we did where it asks us to find the inverse of f(x) = 2^(x+3) ?If so, I can help: When you are finding the inverse, switch the xs and the ys, and solve for the new y. By doing that to our equation, we get: x= 2^(y+3)now we take the log of both sides to bring down the (y+3) exponent:log x= (y+3) log 2now we divide both sides by log 2 to isolate the (y+3) on the right side of the equation:(log x)/(log 2) = (y+3)Notice that the left side of the equation is a change of base deal, so we can turn the equation into:Log base 2 of x = (y+3)All that’s left to do is to subtract the 3:(Log base 2 of x ) – 3 = yAnd don’t forget to write it as y^-1 because that’s how you show it is an inverse. hope this helps!
could someone help me with VI letter k from the review we did in class today?
Can someone please show me how to do 27 on the review?
Shivani:ln y = 2x^2 + ln CTurn 2x^2 into a natural logln y = ln e^(2x^2) + ln CCombine the right sideln y = ln C * e^(2x^2)Simplifyy = ln Ce^(2x^2)Hope this helps!
Justin VI letter k is:log(x)/log(25)+ log(x)/log(5)= 9change of base->(log(x)/log(5))/(log(25)/log(5))+log(x)/log(5)=9simplify denominator->(log(x)/log(5))/(log(5^2)/log(5))+log(x)/log(5)=9(log(5^2)/log(5)) can simplify to 2multiply by 2 ->(log(x)/log(5))+2(log(x)/log(5))=18simplify ->(log(x)/log(5))+(log(x^2)/log(5))=18multiply terms together ->(log(x^3)/log(5))=18bring 3 infront of log->3(log(x)/log(5))=18divide by 3->(log(x)/log(5))=6change to exponential form->5^6=xx=15625
Can someone show me how to do 24 on the review?
Shafia, after much hemming and hawing this is what I got...~ (1/2)ln((x^2)+1)-4ln(1/2)-1/2[ln(x+4)+lnx]~ ln((x^2)+1)^(1/2)-ln(1/2)^4-(1/2)ln(x+4)-(1/2)lnx(distribute the negative)~ ln((x^2)+1)^(1/2) - [ln(1/2)^4 + ln(x-4)^(1/2) + lnx^(1/2)]Take what's in front and make it a power~ ln((x^2)+1)^(1/2) - ln[(1/2)^4 * (x-4)^(1/2) * x^(1/2)]Take the ln out. Now you know it's a difference of two logs, so you need to divide.~ ln [[((x^2)-1)^(1/2))]/[((1/2)^4) * ((x-4)^(1/2)) x (x^(1/2))]]~ I'm not entirely sure we're supposed to stop simplifying, but I went on (just for funsies). You could always also take the square root of (x-4) and x. I did in my homework. I was unable to show the work here because I have no idea how to bring the square root symbol.~ If you were to combine the square root of (x-4) and (x), you would end up with the square root of ((x^2)-4x).~ But I'm sure it's acceptable to stop there.~ However if you continue simplifying, you should get this:(1/2)ln[(256((x^2)-1))/((x^2)-4x)]~ I had a lot of trouble with this one- just because it took me a while to "see" it. Simplifying was a little tedious, too, since there were many places where you could make an error.Hope this helps! :]
Can someone help me with #32 on the review? Much appreciated!
Does anyone remember how to graph a log with a base other than 10 inthe calculator?
Robert, You first have to do the change the base formula so that you have a base 10 log then you can plug it into your claculator and find the graph.
Hey sorry that was me that posted before
number 22 on 5.6?
For 22, I started by setting P and t equal to something. I said P=10,000 and t=1. First, I found the effective rate of the 10,000 dollars. You find this by doing A=Prt, so A=10,700=10,000*.07*1. Next, I found the equation for the compounding quarterly part of the question. By substituting the appropriate numbers into the equation, my equation looked like: A=10,000 (1+ r/4)^(4)(1). Since both the equations equal the variable A, I set them equal to each other: 10,700=10,000 (1+ r/4)^4. I divided by 10,000: 1.07 = (1+ r/4)^4. Then, I took the 4th root of each side: 1.0171 = 1 + r/4. Subtract 1: .0171 = r/4. Multiply by 4: r = 6.82. Hope that helps!
To get the 6.82 at the end I just multiplied the decimal answer by 100 to get the percent rate. Sorry, I like skipping steps!
5.6 #13 what equation do you use to set it up?
Kelly, where did you get the 10,700 from?im confused about that
maddie- for 13 you use the presentvalue formula that says:P=A(1+r/n)^(-n)(t), dont forget the negative!
oh never mind! i figured it out.
for 5.6 can anyone show me how to set up the equations for number 29??
Christine-for 29, set up the equations just like you would normally, but you have to change what it is set equal to to 2P so it would look like:2P=P(1=(.08/12))^12tThen you divide by P and just solve for t.Same thing for the other one, except you apply it to the compounded continuously formula.Could someone help me with number 41 from the same section?
sorry, in the first equation it should be 2P=P(1+(.08/12))^12t
Ok, for #29 from 5.6 I understand how to set it up, but I must be making a mistake in the solving part. I keep getting t=8.28Help?
Lindsey - you have to remember that t is years, so to calculate months you just have to multiply by 12
For 41 you have to use the equation A = P(1 + (r/n))^ntIn the question most of these are givenA = $850000P = $650000t = 3 years n = 1 Because it is annual rater = ?Plug it in and you get850,000 = 650,000 (1 + (r/1)) ^ (1)(3)Then you divide by 650,000 and get1.30769 = (1 + r/1)^3take the cubed root, and (r/1) = r1.0935 = 1+rr=.0935So 9.35%
I'm also really confused on number 29 in tonight's homework. Could someone please hlep me?
Anika: For 29, the question is basically asking for t (time). Your equation would be: 2P= P( 1+ (.08/12)^12t since the question wants to know how much time will pass until your investment is doubled (2P), which is why t is unknown. To solve for t, you first divide both sides by P:2= (1+ (.08/12)^12t. Then, you write this equation in using logs:(log 2)/(log 1+(.08/12))=12t. By dividing both sides by 12, you find that t = 8.7 years. Then, the question asks for t to be found, when the investment is compounded continuously. The equation for this is 2P=Pe^(.o8t). First, divide both sides by P: 2=e^(.08t). Then you find the natural log of both sides:ln2=.08t. By dividing both sides .08, you find that t=8.66 years. Multiply both your answers for t by 12, to find the months. I hope that helps!
on #32, how do you do the first part of the problem, where it is asking the "per annum compounded monthly"? I get the compounded continuously, for the 2nd part, but I'm completely stuck on the first part for that. please help!
oh, and also what is the easiest way to solve #47 in tonight's homework...I tried it but I'm pretty sure I'm doing it the long way, besides the fact that I'm making things way too complicated. help on that would be great as well.
For the per annum compounded monthly thing, you just use the regular present value equation. So your equation is: 175=100(1+.1/12)^12t. When you solve for t (I’m assuming you know how to solve for t), you get t=5.6. I’m doing this without my book and just my spiral, so if I messed up on something let me know.
Hey Chandan,For 32, this is what I did...A = P(1 + r/n)^ntSo set your variables.A = $175P = 100r = .1n = 12t = ? (what you're trying to find out)Per annum means annually.So this would be your equation:175 = 100(1 + (.1/12))^12tSolve from there.I got about 6 years. :]
Ha, Kelly, I like how we post a minute away from each other.
How would you set up number 38 in 5.6? I have it set up as 3000=P(1+.03/n?)^n(.05) trying to find P but I don't know what n is. Is John only paying once for the six months? And would that be bianually?
BFrank, The problem says that it is compounded monthly, so n=12. Can anyone explain how they got the answer for section 5.7 #13? I got k=-.000124 and plugged it in like they did in example 2, but I'm getting 9709.46, which is not correct according to the answers at the back of the book.
For number 10 in section 5.7. Are you supposed to use the uninhibited radiactive decay formula? Because its giving me negative numbers...
Can someone show me how to do number 13 in section 5.7? I'm confused how you set it up.
hey lexy! Make sure u use the radioactive decay formula. This is how i set it up to find k: .5=e^5600k, from there you change it to a natural logs problem and then it looks like ln.5=5600k and from there you can solve for k and it equals around -0.000124. With knowing what k is the next equation looks like 0.3=e^-0.000124t. You then turn this into a natural logs problem and you get ln0.3=-0.000124t and then you can solve for t. I got that it equals around 9709.5 years.
heyy so i was just wondering for #10 in section 5.7 i got the answer of 710,802.61 and the answer key says 711,115 so i was just wondering if im doing this rightt...so i know:t=2A(initial)= 900,000A= 800,000and then I plugged them into the equation:A= A(initial)e^ktand got k= -0.059then, i plugged in 4 for "t" (time) and ended up with my answer: 710,802.61i dont know if im making a mistake in this problem...?
beatrice, your k should equal -.0589
im confused on how you get part c. in the homework for 5.8
I need help with number ten also! It keeps telling me that it can't graph because of the domain! helpp?
For #10 part c, you have to press stat, edit, and make sure you plugged in all the correct values. In order to graph on a clear screen press stat, then go over to cacalc and hit 9 LnReg. Next you have to tell the calculator where to put your values so you do 2nd 1 comma, then 2nd 2 comma, then hit vars go over to y-vars and press 1. So it should read (LnReg L1,L2,Y1) hit enter. then you should able to go to your y= and see the formula there. Next to see just ur scatter points be sure that plot is on and hit zoom and go to stat. The graph should be a curve going down.
Julia: the domain error is most likely caused by an incorrectly defined window screen. You can fix this by hitting "zoom" and selecting ZStandard (6) or ZoomFit (0).
@ Beatrice! KokaRemember to ALWAYS substitute the EXACT value for k.Once you have plugged in everything and derived the value from your calculator, Hit [2ND] [(-)] [STO>] [ALPHA] [(] to store the value into the K variable.You can retrieve the K variable later using [2ND] [(].Yeah, unfortunately one decimal-place off makes a huge difference in this case. :(
For #8 from section 5.8, how did you guys plug the values into L1 and L2 because whenever i try to do it, the calculator keeps on saying, "overflow."
I kept getting that too for #8. What u need to do is substitute the years for smaller digits. So for the first year which is 1989, you change it to 0. You do the same for the rest of the years so L1 will be numbers 0-10. Changing this wont change the problem and now u wont get overflow!
Can someone show me how to do number 4 in section 5.7?
Can anyone show me how to do 19 on 5.7 and or 13 on 5.8?
Can anyone explain when to use the P=e^rt formula? I just dont understand when it use used and would appreciate some clarity
Elayna: to do number 4, plug in values for A(subscript 0) (the initial amount present) and A. A is going to be half of A(subscript 0) because the time you're looking for will decrease the initial amount by half. I plugged in 4 and 2, respectively, and got about 7.97 years as my answer.Hope it cleared things up
Shafia: For 19 on 5.7, you can plug the given values into your calculator by hitting "STAT" under the "DEL" button. After hitting "STAT", press "ENTER" and enter your given x values under L1 and your given y values under L2. After entering those values, hit "STAT" again, press right, and press ExpReg (number 0). This will calculate the equation of the graph.
I have been working on this problem for days...number 22 in section 5.6. anyone know how to do it?
I know this is kind of late, but don't judge!So i am having trouble with radioactive decay problems on how to set them up. Can someone please help by showing how to start number 11 or 12 on page 346 from 5.7?
I don't radioactive decay. Or growth. Or interest. Or cooling.Please help.
I am a little confused about Newton's Law of Cooling, could someone please explain it to me?
Okay Alec. For #11, you'd use the Anod(e^kt) formula. but you'd use the fact that (.5Anod)=Anod(e^k(1690)). Divide by Anod on each side to get .5=e^1690k. when you convert this to exponential form, you get 1690k=ln(.5). divide each side by 1690, and you get k=-4.1*10^-4. Now you can use the actual formula. You would have A(50)=10e^50k. plug everything in and you get 9.797g. And Nithya, the variables in that law are u being the temp. of an obj. and t being elapsed time. the capital T is the temp. of the surrounding medium i think. but its just plugging and chugging for problems with this formula.
can someone show me how to do #22 in section 5.7?
Maddie, For number twenty two, I believe you substitute the information into Newton's law of cooling (even though it's heating). There are two parts of the problem because in order to find out how long you need to keep the pig in the oven, you need to know the constant (k). You can find this by substituting the original information into the law of cooling to get 100=325+(75-325)e^2k. Because its end temperature is one hundred degrees, the temperature of the surrounding medium is 325 degrees (I hope. I'm assuming this because it says the temperature remains at 325?) the temperature of the pig before it was in the over was the 75 degrees, and it was in the oven for two hours. This way, you should get that k is equal to about 1.0986. When you're simplifing you are going to have to convert it into a natural log to find k. Then you substitute the new information into Newton's law of cooling to get 175+325+(100-325)e^1.09861t. And I think you should get about .37 hours.
I am confused on how to start number 13 in 5.7. Thanks in advance!
when trying to "select the best model" for an equation (ExpReg, PwrReg, etc) the "r" doesn't show up on my calculator... is there any other way of solving those types of problems?
sivani, if you scroll up about 25 people the answer is there
Does anyone know if we need to have all the formulas memorized for the test tomorrow or is she going to give them to us?
Elayna - she's going to give them to us
-elaynaMrs. Johnson told us that she will give the formula's on the test, just be sure to know which formula to use on a given problem and you should be fine
For number 13 in 5.7, I get how to solve for the constant K. My question is how do you solve for the time (the second half of the problem)? Someone posted an explanation to this problem earlier and I am confused as to where they got 0.3 in the equation 0.3=e^-0.000124t
On the chapter 5 review can someone help me with the set up for number 71. Much appreciated!
Sorry, disregard my post before this! I was looking at the wrong problem.
For problem 71, you use the Uninhibited Radioactive Decay formula: A(t)=Ae^kt. First, you have to find what k equals. The problem tells you the half-life of carbon-14 is 5600 years. So you set k equal to 5600. Since 5600 is the HALF life, then A(t) must equal a half of A. So you set A(t) equal to .5A. So your new equation to solve for k is: .5A=Ae^5600k. The A’s cancel out through division: .5=e^5600k. Rewrite this in log form: ln .5 = 5600k. Divide by 5600 to isolate k: ln .5/5600 = k. When you solve, you get that k = -.000123776. When you plug this value back into the original formula, you get: A(t) = Ae^-.000123776t. Since the problem gives you that the bones had 5 percent of the original amount of carbon-14, A(t) must equal .05A. So when you plug this back into the equation, you get .05A = Ae^-.000123776t. Now you have to solve for t. Divide by A: .05 = e^-.000123776t. Rewrite in log form: ln .05 = -.000123776t. Divide by -.000123776 to isolate t: ln. 05/-.000123776 = t. When you solve this, you get t = 24203 years.
Thanks! Makes alot more sense now! :)
Could someone show me how to do the half life problems? like number 71 on the review problems.
NEVER MIND, it's right above :)
Does anyone know where to go on part b of number 66 on the review? I got to .98=logd. Thanks.
For 66, at that point, you have to know that log d means log based 10 of d.because it is log base 10 of d, you can bring the 10 to the other side as:10^.98and it should look like:10^.98 = dd = 9.56 inches
Can someone help me with # 73 on the review? I don't know what to do..
I am confused on how to find the half life in problem 4 in 5.8
Does anyone know how to set up #74 on the review, page 364? I'm a little confused on that one. Thanks!
How do you solve question # 72 in the review. I'm using the formula for newtons law of cooling, but I'm not getting the right answer. Can somebody show how to do this problem.
What is the key in the zoom menuthat you use to fit the scatter diagrams in the window? Is it ZoomStat?
Mrs. Johnson teaches math at Metea Valley High School in Aurora, IL.
Ap Ab Calculus
H Algebra 2 Trig
H Pre Calculus
Regular Pre Calculus