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 8.4b:

$\log _{12}\left( 4x-5\right) ^{2}=16$


Answer:

x=107,495,425.25 and $107,495,422.75\bigskip
\bigskip $


Solution:



The above equation is valid only if the term $\quad \log _{12}\left(
4x-5\right) ^{2}$ is valid. The term $\log _{12}\left( 4x-5\right) ^{2}$ is valid if $\left( 4x-5\right) ^{2}>0\longrightarrow x\neq \displaystyle \frac{5}{4}.$Therefore, the equation is valid when $x\neq \displaystyle \frac{5}{4}.$ Another way of saying this is that the domain is the set of real numbers where $x\neq \displaystyle \frac{5}{4}.$


Convert the logarithmic equation to an exponential equation with base 12.


\begin{eqnarray*}&& \\
\log _{12}\left( 4x-5\right) ^{2} &=&16 \\
&& \\
&& \\
12^{16} &=&\left( 4x-5\right) ^{2} \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
\pm 12^{8} &=&4x-5 \\
&& \\
&& \\
5\pm 12^{8} &=&4x \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
\displaystyle \frac{5\pm 12^{8}}{4} &=&\displaystyle \fr...
...& \\
&& \\
x &=&\displaystyle \frac{5\pm 12^{8}}{4} \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
x &=&\displaystyle \frac{5+12^{8}}{4}=107,495,425.25 \\ ...
...\
x &=&\displaystyle \frac{5-12^{8}}{4}=-107,495,422.75 \\
&&
\end{eqnarray*}


The exact answers are x=107,495,425.25 and x=-107,495,422.75. These answers may or may not be solutions to the original equation. You must check them in the original equation, either by numerical substitution or by graphing.


Numerical Check:



Left Side: $\qquad \log _{12}\left( 4x-5\right) ^{2}=\log _{12}\left( 4\left(
107,495,425.25\right) -5\right) ^{2}$

\begin{eqnarray*}&& \\
&=&\log _{12}\left( 4\left( 107,495,425.25\right) -5\rig...
... 10^{17}\right) }{\ln \left(
12\right) } \\
&& \\
&=&16 \\
&&
\end{eqnarray*}


Right Side:$\qquad 16$



Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 107,495,425.25 for x, then x=107,495,425.25 is a solution.





Left Side: $\qquad \log _{12}\left( 4x-5\right) ^{2}=\log _{12}\left( 4\left(
-107,495,422.75\right) -5\right) ^{2}$
\begin{eqnarray*}&& \\
&=&\log _{12}\left( 4\left( -107,495,422.75\right) -5\ri...
... 10^{17}\right) }{\ln \left(
12\right) } \\
&& \\
&=&16 \\
&&
\end{eqnarray*}


Right Side:$\qquad 16$



Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 1107,495,422.75 for x, then x=-107,495,422.75 is a solution.





Note: If you had simplified the problem to $\log _{12}\left( 4x-5\right) =8,$you would have lost one of the answers.



Graphical Check:


You can also check your answer by graphing $\quad f(x)=\log _{12}\left(
4x-5\right) ^{2}-16\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 107,495,425.25 and -107,495,422.75. This means that 107,495,425.25 and -107,495,422.75 are the real solutions.



If you have trouble graphing the function $f(x)=\log _{12}\left( 4x-5\right)
^{2}-16\quad $, graph the equivalent function $f(x)=\displaystyle \frac{\ln 4x-5^{2}}{\ln
\left( 12\right) }-16\quad $





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