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 8.4d:

$\log _{.982}\left( x^{2}-6x+5\right) ^{2}=8$


Answer:

$x=3\pm \sqrt{4+\left(
0.982\right) ^{4}}$ and $x=3\pm \sqrt{4-\left( 0.982\right) ^{4}}.$ The approximate answers are $x\approx 0.779657509082,$ 1.2478358458, 4.75216415412, and $5.22034249092.\bigskip\bigskip $



Solution:


The above equation is valid only if the expression $\quad \left(
x^{2}-6x+5\right) ^{2}>0.$ The expression $\left( x^{2}-6x+5\right) ^{2}>0$is valid when $x\neq 1$ and $x\neq 5.$Another way of saying this is that the domain is the set of real numbers where $x\neq 1\ or\ 5.$


Note: If, when solving the above problem, you simplify the logarithmic equation $\log _{.982}\left( x^{2}-6x+5\right) ^{2}=8$ to $\log
_{.982}\left( x^{2}-6x+5\right) =4,$ you will lose half of your solutions. Why?


Recall that $\log _{.982}\left( x^{2}-6x+5\right) ^{2}$ is valid for all real numbers that are not equal to 1 or 5. However, $2\log _{.982}\left(
x^{2}-6x+5\right) $ is valid for the set of real numbers $\left( -\infty
,1\right) \cup \left( 5,\infty \right) .$ This means that $\log
_{.982}\left( x^{2}-6x+5\right) ^{2}=2\log _{.982}\left( x^{2}-6x+5\right) $only over the set of real numbers $\left( -\infty ,1\right) \cup \left(
5,\infty \right) .\bigskip $


So be careful when you simplify a logarithmic equation before solving it.



Convert the logarithmic equation $\log _{.982}\left( x^{2}-6x+5\right) ^{2}=8$ to an equivalent exponential equation with base 0.982.

\begin{eqnarray*}&& \\
\left( 0.982\right) ^{8} &=&\left( x^{2}-6x+5\right) ^{2...
...ft( 0.982\right) ^{4} &=&\pm \left( x^{2}-6x+5\right) \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
x^{2}-6x+5 &=&\pm \left( 0.982\right) ^{4} \\
&& \\
&& \\
x^{2}-6x+5+4 &=&\pm \left( 0.982\right) ^{4}+4 \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
\left( x-3\right) ^{2} &=&\pm \left( 0.982\right) ^{4}+4...
...& \\
x-3 &=&\pm \sqrt{4\pm \left( 0.982\right) ^{4}} \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
x &=&3\pm \sqrt{4\pm \left( 0.982\right) ^{4}} \\
&&
\end{eqnarray*}



\begin{eqnarray*}x &=&3+\sqrt{4+\left( 0.982\right) ^{4}}\approx 5.22034249092 \...
...sqrt{4-\left( 0.982\right) ^{4}}\approx 4.75216415413 \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
x &=&3-\sqrt{4+\left( 0.982\right) ^{4}}\approx 0.779657...
...3-\sqrt{4-\left( 0.982\right) ^{4}}\approx 1.24783584587 \\
&&
\end{eqnarray*}


The exact answers are $x=3+\sqrt{4+\left( 0.982\right) ^{4}},\ x=3+\sqrt{
4-\left( 0.982\right) ^{4}},...
...-\sqrt{4+\left( 0.982\right) ^{4}},\ and\
x=3-\sqrt{4-\left( 0.982\right) ^{4}}$. The approximate answers are $
x\approx 5.22034249092,\ x\approx 4.75216415413,\ x\approx 0.779657509082,\
and\ x\approx 1.24783584587.$



These answers may or may not be the solutions. You must check them numerically or graphically.


Numerical Check:


All four of the answers are the solutions to the original equation.



Graphical Check:


You can also check your answer by graphing $\quad f(x)=\log _{.982}\left(
x^{2}-6x+5\right) ^{2}-8\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 four places: $0.779657509082,\ 1.24783584587,\ 4.75216415413$, and 5.22034249092. This means that $0.779657509082,\ 1.24783584587,\ 4.75216415413$, and 5.22034249092 are the real solutions.



If you have trouble graphing the function $f(x)=\log _{.982}\left(
x^{2}-6x+5\right) ^{2}-8\quad $, graph the equivalent function $f(x)=\displaystyle \frac{
\ln \left( x^{2}-6x+5\right) ^{2}}{\ln \left( 0.982\right) }-8\quad $.





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