Solving an equation with exponents by using logarithms Solve the equation
$$0.25^5 = 4^{(5x-3)/3} \cdot (0.125)^{6x}$$
So would I just bring down the exponents by taking the log of each constant?

Logarithms here are actually unnecessary. Notice that each base is a power of $2$, so we get:
\begin{align*}
0.25^5 &= 4^{\frac{1}{3}(5x - 3)} \cdot (0.125)^{6x} \\
(2^{-2})^5 &= (2^2)^{\frac{1}{3}(5x - 3)} \cdot (2^{-3})^{6x} \\
2^{-10} &= 2^{\frac{2}{3}(5x - 3)} \cdot 2^{-18x} \\
2^{-10} &= 2^{\frac{2}{3}(5x - 3) - 18x} \\
\end{align*}
Since the bases are the same, we may equate exponents:
\begin{align*}
-10 &= \frac{2}{3}(5x - 3) - 18x \\
-30 &= 2(5x - 3) - 54x \\
-30 &= (10x - 6) - 54x \\
-24 &= -44x \\
x &= 6/11
\end{align*}
