joan took out a mortgage from hel local bank. each monthly mortgage payment she makes must be triple the amount of the previous month ' s payment. if her first payment is $ 100, and the total amount she must pay back is $ 36400, how many months will it take joan to pay back her mortgage?  Choose the correct option:
A) 6
B) 8
C) 10
D) 11
E) 13
<gadget id="calculator">36_400 * 2</gadget>
<output>72_800</output>

<gadget id="calculator">72_800 / 100</gadget>
<output>728</output>

<gadget id="calculator">728 + 1</gadget>
<output>729</output>

<gadget id="calculator">log(729)</gadget>
<output>log(729) = around 6.591674</output>

<gadget id="calculator">log(3)</gadget>
<output>log(3) = around 1.098612</output>

<gadget id="calculator">log(729) / log(3)</gadget>
<output>log(729)/log(3) = around 6</output>

<result>A</result>
joan took out a mortgage from hel local bank. each monthly mortgage payment she makes must be triple the amount of the previous month ' s payment. if her first payment is $ 100, and the total amount she must pay back is $ 36400, how many months will it take joan to pay back her mortgage?
divide(log(add(divide(multiply(36400, const_2), 100), const_1)), log(const_3))
"joan starts off with 100 $ . . which is to be tripled every month her monthly payments look like this : 100 , 300 , 900 , 2700 . . . . . . . . . upto 36400 this can be re written as : 100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 . . . . . . 100 x 364 so we have 1 , 3 , 9 , 27 . . . . . 36400 in gp we know that a = 1 , and r = 3 ( its easy to figure it out by looking at the question , but regardless of it being mentioned in the question we can still compute the value of r using the formula tn = a 3 ^ n - 1 . . . ) therefore to find the sum of n terms of a gp we use this formula : sn = a ( 1 - r ^ n ) / 1 - r using this and plugging in the information we get . . . 364 = 1 - 3 ^ n / 1 - 3 ; 1 - 3 ^ n / - 2 cross multiplying we get 364 x - 2 = 1 - 3 ^ n - 728 = 1 - 3 ^ n - 729 = - 3 ^ n 729 = 3 ^ n ( negatives cancel out ) 729 can also be re written as 3 ^ 6 therefore ; 3 ^ 6 = 3 ^ n thus n = 6 ( a )"