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}. $




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