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 function.



Solve for x in the following equation.


Example 3:

$\log _{8}\left( 7x+5\right) -\log _{8}\left( x-10\right)
=\log _{8}\left( x+2\right) $

The above equation is valid only if all the terms in the equation are valid. The term $\log _{8}\left( 7x+5\right) $ is valid if $\left( 7x+5\right)
>0\longrightarrow x>-\displaystyle \frac{5}{7};$ the term $\log _{8}\left( x+5\right) $is valid if $x-10>0\longrightarrow x>10;$ and the term $\log _{8}\left(
x\right) $ is valid if $x+2>0\rightarrow x>-2.$ The domain is the set of real numbers that are greater than $-\displaystyle \frac{5}{7}$, greater than 10, and greater than -2. The domain is therefore the set of real numbers greater than 10.


Algebraic Method:

\begin{eqnarray*}&& \\
\log _{8}\left( 7x+5\right) -\log _{8}\left( x-10\right)...
...( 7x+5\right) }{\left( x-10\right) } &=&\left( x+2\right) \\
&&
\end{eqnarray*}
\begin{eqnarray*}&& \\
\left( x-10\right) \displaystyle \frac{\left( 7x+5\right...
...) \left( x+2\right) \\
&& \\
&& \\
7x+5 &=&x^{2}-8x-20 \\
&&
\end{eqnarray*}
\begin{eqnarray*}&& \\
0 &=&x^{2}-15x-25 \\
&& \\
&& \\
x &=&\displaystyle \frac{15\pm \sqrt{325}}{2} \\
&&
\end{eqnarray*}
\begin{eqnarray*}&& \\
x &=&\displaystyle \frac{15+\sqrt{325}}{2}\approx 16.513...
...15-\sqrt{325}}{2}\approx -1.51387818866> 10 \\
&& \\
&& \\
&&
\end{eqnarray*}


Although there are two solutions, only one of the solutions is in the domain of the problem. The exact solution to the problem is $x=\displaystyle \frac{15+\sqrt{325}
}{2}$, and the approximate solution is $x\approx 16.5138781887.$


Graphing Method:

Change the original equation into an equation where all the logarithmic terms have base 10.

\begin{eqnarray*}&& \\
\log _{6}\left( 2x-3\right) +\log _{6}\left( x+5\right) ...
...og \left( x\right) }{\log
\left( 3\right) } \\
&& \\
&& \\
&&
\end{eqnarray*}


Rewrite the equation as $\displaystyle \frac{\log \left( 2x-3\right) }{\log \left(
6\right) }+\displayst...
...right) }-\displaystyle \frac{
\log \left( x\right) }{\log \left( 3\right) }=0. $

Let's call the left side of the equation f(x) and the right side of the equation g(x).

Then $f\left( x\right) =\displaystyle \frac{\log \left( 2x-3\right) }{\log \left(
6\r...
...( 6\right) }-\displaystyle \frac{
\log \left( x\right) }{\log \left( 3\right) }$ and $g\left( x\right) =0.$Graph f(x) and g(x). We are looking for the point(s) of intersection, $
\left( x,y\right) .$ The solution, if any, will be the value of x in the point(s) of intersection.

The graph of the right side of the equation is the set of points where the value of y equals zero. This is easy; it is the x-axis. We then look to see where the graph of f(x) crosses the x-axis.

Note that the graph only appears to the right of $x=\displaystyle \frac{3}{2}.$ This is consistent with our finding that the domain of the original equation is the set of real numbers greater than $\displaystyle \frac{3}{2}$.

The solution(s) to the original equation is the set of real numbers where f(x) crosses the x-axis (the x-intercepts are the solutions to the problem.) . You will note from the graph that f(x) crosses the x-axis at x=1.6735161761. This means that the equation has one real solution at $
x\approx 1.6735161761 $ .


Algebraic Method:

Change the bases of the logarithmic terms to 10.

\begin{eqnarray*}&& \\
\log _{6}\left( 2x-3\right) +\log _{6}\left( x+5\right) ...
...og \left( x\right) }{\log
\left( 3\right) } \\
&& \\
&& \\
&&
\end{eqnarray*}


Multiply both sides of the equation by the factor $\left( \log \left(
6\right) \right) \left( \log \left( 3\right) \right) .$

\begin{eqnarray*}&& \\
&& \\
\left( \log \left( 6\right) \right) \left( \log \...
...tyle \frac{\log \left( x\right)
}{\log \left( 3\right) } \\
&&
\end{eqnarray*}
\begin{eqnarray*}&& \\
\log \left( 3\right) \cdot \log \left( 2x-3\right) +\log...
...right) } &=&\log \left( x\right) ^{\log \left( 6\right) } \\
&&
\end{eqnarray*}
\begin{eqnarray*}&& \\
\log \left( 2x-3\right) ^{\log \left( 3\right) }\cdot \l...
...&=&\left( x\right) ^{\log \left( 6\right) } \\
&& \\
&& \\
&&
\end{eqnarray*}
An algebraic solution, other than interpolation, is too difficult for a beginning student.



Interpolation Method:

Since the domain is the set of real numbers greater than $\displaystyle \frac{3}{2},$choose numbers larger than $\displaystyle \frac{3}{2}. $




If you would like to test yourself by working some problems similar to this example, click on problem.


If you would like to go to the next section, click on next.


If you would like to go back to the previous section, click on previous.


If you would like to go back to the equation table of contents, click on contents.


This site was built to accommodate the needs of students. The topics and problems are what students ask for. We ask students to help in the editing so that future viewers will access a cleaner site. If you feel that some of the material in this section is ambiguous or needs more clarification, or you find a mistake, please let us know by e-mail.



[Algebra] [Trigonometry]
[Geometry] [Differential Equations]
[Calculus] [Complex Variables] [Matrix Algebra]

S.O.S. MATHematics home page


Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.


Author: Nancy Marcus

Copyright 1999-2017 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour