Question 881970
I can't solve this problem. Please help me. 
(3^x)(9^x-1) = 27^1-x

My solution: 
27^2x-1 = 27^1-x
I don't know what to do next. Thank You. 
<pre>
{{{(3^x)(9^(x - 1)) = 27^(1 - x)}}}
{{{(3^x)(3^2)^(x - 1) = (3^3)^(1 - x)}}} ------ Converting 9 and 27 to base 3 
{{{(3^x)(3^(2x - 2)) = 3^(3 - 3x)}}}
{{{3^(x + 2x - 2) = 3^(3 - 3x)}}}
x + 2x - 2 = 3 - 3x ------ Bases are equal and so are their EXPONENTS
3x + 3x = 3 + 2
6x = 5
{{{highlight_green(highlight_green(x = 5/6))}}}