Question 983070
<pre>
I won't do your problem for you.  Instead here is one EXACTLY like yours.
Use it as a model to solve yours:

{{{(1/27)^(3x-4)=9^(x-5)}}}

Write 27 as 3³ and 9 as (3²)

{{{(1/3^3)^(3x-4)=(3^2)^(x-5)}}}

Write {{{1/3^3}}} as {{{3^(-3)}}}

{{{(3^(-3))^(3x-4)=(3^2)^(x-5)}}}

remove the parentheses by multiplying the exponents:

{{{3^(-9x+12)=3^(2x-10)}}}

Since the bases are the same, and they are positive
and not equal to 1, we may equate the exponents only

{{{-9x+12=2x-10}}}

{{{-9x+12=2x-10}}}

{{{-11x=-22}}}

{{{x=2}}}

Edwin</pre>