College Algebra, Chapter 5, 5.4, Section 5.4, Problem 22

Determine the solution of the exponential equation $10^{1-x} = 6^x$.

$
\begin{equation}
\begin{aligned}
10^{1-x} &= 6^x\\
\\
\log 10^{1-x} &= \log 6^x && \text{Take $\log$ of each side}\\
\\
(1-x)\log 10 &= x \log 6 && \text{Law of Logarithms } \log_a A^c = C \log_a A\\
\\
1 - x &= \frac{x \log 6}{\log 10} && \text{Divide by } \log 10\\
\\
1 &= x \frac{\log 6}{\log 10 } + x && \text{ Add } x\\
\\
1 &= x \left( \frac{\log 6}{\log 10} + 1 \right) && \text{Factor out } x\\
\\
x &= \frac{1}{\left( \frac{\log 6}{\log 10} + 1 \right)} && \text{Divide by } \left( \frac{\log 6}{\log 10} + 1 \right)\\
\\
x &= 0.5624
\end{aligned}
\end{equation}
$