Solving Logarithmic Equations

Solve each equation:

1. \(\log _{3}( 4x-7) =2\)
2. \(\log _{x}64=2\)

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 \(8\).