If you would like an in-depth review of logarithms, the rules of logarithms, logarithmic functions and logarithmic equations, click on logarithmic functions.

Solve for x in the following equation.

Problem 8.4c:

$\log _{\displaystyle \frac{4}{3}}\left( 6x-11\right) ^{3}=5$


$x=\displaystyle \frac{\left( \displaystyle \frac{4}{3}
\right) ^{\displaystyle \frac{5}{3}}+11}{6}.$ The approximate answer is $x\approx
2.10253638412\bigskip\bigskip $


The above equation is valid only if the term $\quad \log _{\displaystyle \frac{4}{3}
}\left( 6x-11\right) ^{3}$ is valid. The term $\log _{\displaystyle \frac{4}{3}}\left(
6x-11\right) ^{3}$ is valid only if $\left( 6x-11\right) ^{3}>0\longrightarrow x>
\displaystyle \frac{11}{6}.$ Therefore, the equation is valid when $x>\displaystyle \frac{11}{6}.$Another way of saying this is that the domain is the set of real numbers where $x>\displaystyle \frac{11}{6}.$

Convert the logarithmic equation to an exponential equation with base $
\displaystyle \frac{4}{3}.$

\begin{eqnarray*}&& \\
\log _{\displaystyle \frac{4}{3}}\left( 6x-11\right) ^{3...
...e \frac{4}{3}\right) ^{5} &=&\left( 6x-11\right) ^{3} \\
&& \\
\begin{eqnarray*}&& \\
\left[ \left( \displaystyle \frac{4}{3}\right) ^{5}\righ...{4}{3}\right) ^{\displaystyle \frac{5}{3}} &=&6x-11 \\
&& \\
\begin{eqnarray*}&& \\
\left( \displaystyle \frac{4}{3}\right) ^{\displaystyle ...
...yle \frac{5}{3}}+11}{6} &=&\displaystyle \frac{6x}{6} \\
&& \\
\begin{eqnarray*}&& \\
x &=&\displaystyle \frac{\left( \displaystyle \frac{4}{3...
... ^{\displaystyle \frac{5}{3}}+11}{6}\approx
2.10253638412 \\

The exact answer is $x=\displaystyle \frac{\left( \displaystyle \frac{4}{3}\right) ^{\displaystyle \frac{5}{3}}+11}{6}
;$ the approximate answer is $x\approx 2.10253638412.$ This answer may or may not be the solution to the original equation. You must check the answer in the original equation, either by numerical substitution or by graphing.

Numerical Check:

Left Side: $\qquad \log _{\displaystyle \frac{4}{3}}\left( 6x-11\right) ^{3}$

\begin{eqnarray*}&& \\
&=&\log _{\displaystyle \frac{4}{3}}\left( 6\left( 2.102...
...\left( \displaystyle \frac{4}{3}\right) } \\
&& \\
&=&5 \\

Right Side:$\qquad 5$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 2.10253638412 for x, then x=2.10253638412 is a solution.

Graphical Check:

You can also check your answer by graphing $\quad f(x)=\log _{\displaystyle \frac{4}{3}
}\left( 6x-11\right) ^{3}-5\quad $(formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; that will be the real solution. Note that the graph crosses the x-axis at 2.10253638412. This means that 2.10253638412 is the real solution.

If you have trouble graphing the function $f(x)=\log _{\displaystyle \frac{4}{3}}\left(
6x-11\right) ^{3}-5\quad $, graph the equivalent function $f(x)=\displaystyle \frac{\ln
\left( 6x-11\right) ^{3}}{\ln \left( \displaystyle \frac{4}{3}\right) }-5\quad $

If you would like to review the solution to problem 8.4d, click on solution.

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