College Algebra, Chapter 5, 5.4, Section 5.4, Problem 46

Solve the Logarithmic Equation $2 \log x = \log 2 + \log (3x - 4)$ for $x$.

$
\begin{equation}
\begin{aligned}
2 \log x &= \log 2 + \log (3x - 4)\\
\\
\log x ^2 &=\log 2 + \log(3x - 4) && \text{Law of Logarithm } \log_a A^c = C \log_a A\\
\\
10^{\log x^2} &= 10^{\log 2} + 10^{\log (3x - 4)} && \text{Raise 10 to each side}\\
\\
x^2 &= 2 + 3x - 4 && \text{Property of } \log\\
\\
x^2 - 3x + 2 &= 0 \\
\\
(x - 2) (x - 1) &= 0
\end{aligned}
\end{equation}
$

Solve for $x$

$
\begin{equation}
\begin{aligned}
x - 2 &= 0 &&\text{and}& x -1 &= 0 \\
\\
x &= 2 &&& x &= 1
\end{aligned}
\end{equation}
$

The only solution in the given equation is $x = 2$, since $x = 1$ will make the term $\log(3x - 4)$ negative.