Solve each equation:

Solution (a) We change the logarithmic equation to an exponential equation. $$\begin{eqnarray*} \log _{3}( 4x-7)~&=&2 \\[3pt] 4x-7 &=&3^{2}{\rm \ \ \ \ \ \ \ \ {\color{#0066A7}{{\hbox{Change to an exponential equation.}}}}} \\[3pt] 4x-7 &=&9 \\[3pt] 4x &=&16 \\[3pt] x &=&4 \end{eqnarray*}$$

Check: For $x=4$, $\log _{3}( 4x-7) =\log _{3}( 4\cdot 4-7) =\log _{3}9=2$, since $3^{2}=9$.

The solution is $4$.

(b) We change the logarithmic equation to an exponential equation. $$\begin{array}{rcl@{\quad\qquad}l} \log _{x}64 &=&2 \\[3pt] x^{2} &=&64 & {\color{#0066A7}{{\hbox{Change to an exponential equation.}}}} \\[3pt] x &=& 8 \quad \hbox{or}\quad x=-8 & {\color{#0066A7}{{\hbox{Solve.}}}} \end{array}$$

The base of a logarithm is always positive. As a result, we discard $-8$ and check the solution $8$.

Check: For $x=8$, $\log _{8} 64=2$, since $8^{2}=64.$

The solution is $8$.