laws of exponents?!?

Question

ha ha i don't know why i'm always so bad at these. anyway, umm here's the problem:

(3*5^X-5) / (27*5^x-3)

how do i solve this for x? i know there's another way...but usually i would just set 27 as a power of 3 and then easily subtract the exponents using the quotient rule. however, this would cancel out the Xs, thus, i am a little confused on how to go about this. i know the right answer is 1/225 thanks to my teacher's key, but HOW? (this is on a quiz review)

Answer 1

3*5^(x - 5)/27*5^(x - 3) =

(3/27)[5^(x - 5)/5^(x-3) =

(1/9)[5^[x-5 -(x - 3)] (since a^x/a^y = a^(x - y))

(1/9[5^(x - 5 - x + 3) =

(1/9)[5^(-2) =

(1/9)(1/5^2) =

1/(9*25) =

1/ 225

Answer 2

Hello,

take a 3 out of the top and bottom giving us 5^(x-5)/9^(5x-3) now subtract the exponents on the 5 giving us 5^-2/9 or 1/9*(5^2) or 1/9*25 giving us 1/225.

Hope This Helps!!