WE WANT TO WRITE THE 
LOGARITHMIC EQUATIONS
AS EXPONENTIAL EQUATIONS.
TO DO THIS, WE NEED TO 
UNDERSTAND THIS LOG EQUATION
AND THIS EXPONENTIAL EQUATION 
MEAN THE SAME THING
OR EQUIVALENT.
LOG BASE E OF N = A MEANS 
B RAISED TO THE POWER OF A = N.
SO B IS THE BASE IN BOTH 
LOG FORM AND EXPONENTIAL FORM.
A IS THE EXPONENT BOTH IN 
LOG FORM AND EXPONENTIAL FORM.
AND N IS THE NUMBER IN BOTH 
LOG FORM AND EXPONENTIAL FORM.
SO REMEMBER A LOGARITHM 
IS AN EXPONENT
AND THAT'S WHY IT'S EQUAL TO A,
WHICH IS THE EXPONENT 
IN EXPONENTIAL FORM.
SO TO WRITE LOG BASE 3 OF 81 = 4 
IN EXPONENTIAL FORM,
WE NEED TO RECOGNIZE THAT 
3 IS THE BASE.
SINCE A LOGARITHM IS AN 
EXPONENT, 4 IS THE EXPONENT,
AND THIS MUST EQUAL THE NUMBER, 
WHICH IS 81.
WHEN CONVERTING FROM LOG FORM 
TO EXPONENTIAL FORM,
WE CAN VERIFY THIS IS CORRECT BY 
JUST EVALUATING THE LEFT SIDE
AND MAKING SURE IT EQUALS 
THE RIGHT SIDE.
3 TO THE FOURTH DOES = 81, 
THERE FORE THIS IS CORRECT.
ONE MORE WAY TO DO THIS 
IS TO START WITH THE BASE,
WHICH IN THIS CASE WOULD BE 3,
AND WORK OUR WAY 
AROUND THE EQUAL SIGN
TO FORM OUR EXPONENTIAL 
EQUATION.
NOTICE 3 TO THE FOURTH IS = 
TO 81.
NEXT WE HAVE LOG BASE 2 OF 64 
= 6, SO OUR BASE IS 2,
THE EXPONENT IS 6, 
AND THE NUMBER IS 64.
2 TO THE SIXTH = 64, 
WHICH IS TRUE.
OR, AGAIN, WE CAN START WITH 2,
WORK OUR WAY AROUND THE EQUAL 
SIGN, 2 TO THE SIXTH = 64.
NEXT WE HAVE LOG BASE 81 OF 9 
= 1/2,
SO OUR BASE THIS TIME IS 81, 
THE EXPONENT IS 1/2,
AND THE NUMBER IS 9.
REMEMBER RAISING SOMETHING 
TO THE 1/2 POWER
IS THE SAME AS 
TAKING THE SQUARE ROOT.
AND SINCE THE SQUARE ROOT OF 81 
= 9 THIS CHECKS.
AND, AGAIN, 
81 TO THE 1/2 POWER = 9.
FOR OUR LAST EXAMPLE 
WE HAVE LOG BASE 5 OF 1/25 = -2,
SO OUR BASE IS 5, THE EXPONENT 
IS -2, AND THE NUMBER IS 1/25,
WHICH AGAIN IS CORRECT.
REMEMBER 5 TO THE -2 
IS = TO 1/5 SQUARED,
WHICH DOES = 1/25.
SO 5 TO THE -2 IS = TO 1/25.
NOW, I DO WANT TO MENTION 
ONE LAST THING.
FOR THESE EXAMPLES I'VE BEEN 
SAYING LOG BASE B OF N = A,
BUT YOU'LL ALSO OFTEN HEAR 
LOG N BASE B = A,
WHICH WOULD BE EQUIVALENT.
OKAY, I HOPE THIS HELPS.
 
