SOLVING LOGARITHMIC EQUATIONS


Note:

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 8.4a:

$\log _{8}\left( 3x+4\right) ^{4}=24$


Answers:

x=87,380 and $x=-\displaystyle \frac{262,148}{3}\bigskip
\bigskip $


Solution:



The above equation is valid only if $\quad \log _{8}\left( 3x+4\right) ^{4}$is valid. The term $\log _{8}\left( 3x+4\right) ^{4}$ is valid if $\left(
3x+4\right) ^{4}>0\longrightarrow x\neq -\displaystyle \frac{4}{3}.$ Therefore, the equation is valid when $x\neq -\displaystyle \frac{4}{3}.$ Another way of saying this is that the domain is the set of real numbers where $x\neq -\displaystyle \frac{4}{3}.$



If you choose to work the problem by first removing the exponent 4, you will lose one of the solutions because $\log _{8}\left( 3x+4\right) ^{4}$ is equivalent to $4\log _{8}\left( 3x+4\right) $ only when $x>-\displaystyle \frac{4}{3}$.


Covert the logarithmic equation to an exponential equation.


\begin{eqnarray*}&& \\
\log _{8}\left( 3x+4\right) ^{4} &=&24 \\
&& \\
&& \\
8^{24} &=&\left( 3x+4\right) ^{4} \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
\pm 8^{6} &=&3x+4 \\
&& \\
&& \\
\pm 8^{6}-4 &=&3x \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
\displaystyle \frac{\pm 8^{6}-4}{3} &=&\displaystyle \fr...
...& \\
&& \\
x &=&\displaystyle \frac{\pm 8^{6}-4}{3} \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
x &=&\displaystyle \frac{8^{6}-4}{3}=87,380 \\
&& \\
&...
...ystyle \frac{262,148}{3}\approx -87,382.6666666667 \\
&& \\
&&
\end{eqnarray*}

The exact answers are $x=\displaystyle \frac{\pm 8^{6}-4}{3}.\bigskip\bigskip\bigskip $

These answers may or may not be the solutions to the original equation. You must check them in the original equation, either by numerical substitution or by graphing.

Numerical Check:

Left Side:

$\qquad \log _{8}\left( 3x+4\right) ^{4}=\log _{8}\left( 3\left(
87,380\right) +4\right) ^{4}$

\begin{eqnarray*}&& \\
&=&\log _{8}\left( 4.72236648287\times 10^{21}\right) \\...
... 10^{21}\right) }{\ln \left( 8\right)
} \\
&& \\
&=&24 \\
&&
\end{eqnarray*}


Right Side:$\qquad 24$



Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 87,380 for x, then x=87,380 is a solution.





Left Side: $\qquad \log _{8}\left( 3x+4\right) ^{4}=\log _{8}\left( 3\left( -
\displaystyle \frac{262,148}{3}\right) +4\right) ^{4}$

\begin{eqnarray*}&& \\
&=&\log _{8}\left( 4.72236648287\times 10^{21}\right) \\...
... 10^{21}\right) }{\ln \left( 8\right)
} \\
&& \\
&=&24 \\
&&
\end{eqnarray*}

Right Side:$\qquad 24$



Since the left side of the original equation is equal to the right side of the original equation after we substitute the value $-\displaystyle \frac{
262,148}{3}$for x, then $x=-\displaystyle \frac{262,148}{3}$ is a solution.





Graphical Check:


You can also check your answer by graphing $\quad f(x)=\log _{8}\left(
3x+4\right) ^{4}-24\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 87,380 and -87,382.66666667. This means that 87,380 and -87,382.66666667 are the real solutions.


If you have trouble graphing the function $f(x)=\log _{8}\left( 3x+4\right)
^{4}-24$, graph the equivalent function $f(x)=\displaystyle \frac{\ln \left( 3x+4\right)
^{4}}{\ln \left( 8\right) }-24$.





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


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