# Thread: exponential problem

1. ## exponential problem

Solve the following for x.

2. $\begin{gathered}
3^{4x} 3^{ - x - 2} = 3^3 3^{ - 2x} \hfill \\
3^{3x - 2} = 3^{3 - 2x} \hfill \\
\Leftrightarrow 3x - 2 = 3 - 2x \hfill \\
\Leftrightarrow x = 1 \hfill \\
\end{gathered}
$

3. ## Solution

I think I got it:

Rewrite all of the terms as base 3:

9^(2x) = (3^2)^(2x) = 3^(4x)
(1/3)^(x+2) = 3^(-x-2)
27 = 3^3
(3^x)^(-2) = 3^(-2x)

Now rewrite the original equation:
(3^(4x)) * (3^(-x-2)) = (3^3) * 3^(-2x)

Now simplify by adding exponents as follows:
3^(3x-2) = 3^(3-2x)

Now, take the log of both sides to yield:

(3x-2)*log3 = (3-2x)*log3

Cancel log 3 from both sides. You are left with:
3x-2 = 3-2x
5x=5
x=1

If you substitute x = 1 back in it should check out.