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 8.5c:

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


Answers: There is one answer. The exact answer is $x=\displaystyle \frac{ 15+\sqrt{325}}{2}$, and the approximate answer is $x\approx 16.513878$.


Solution:

The above equation is valid only if each of the three terms is valid. The term $\log _{8}\left( 7x+5\right) $ is valid if $7x+5>0\rightarrow x>-\displaystyle \frac{ 5}{7}.$ The term $\log _{8}\left( x-10\right) $ is valid if $ x-10>0\rightarrow x>10.$ The term $\log _{8}\left( x+2\right) $ is valid if $ x+2>0\rightarrow x>-2.$ Therefore, the equation is valid when the domain is the set of real numbers is greater than $-\displaystyle \frac{5}{7}$, greater than 10, and greater than -2. This means that the equation is valid if we restrict the domain to the set of real numbers greater than 10.



Simplify the equation and solve.

\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*}&& \\ 7x+5 &=&\left( x-10\right) \left( x+2\right) \\ && \\ ... ...-4\left( 1\right) \left( -25\right) }}{2\left( 1\right) } \\ && \end{eqnarray*}
\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{15\pm \sqrt{325}}{2} \\ && \\ ... ...yle \frac{15-\sqrt{325}}{2}\approx -1.513878\\ && \\ && \\ && \end{eqnarray*}

There are two answers; however, only one of them is greater than 10. This answer may or may not be a solution. You must check it with the original equation either numerically or graphically.




Suppose you did not go through the initial exercise and wanted to check both answers. You can check them in the original equation, either by numerical substitution or by graphing.

Numerical Check:

Left Side: $\qquad \log _{8}\left( 7x+5\right) -\log _{8}\left( x-10\right) $

\begin{eqnarray*}&& \\ &=&\log _{8}\left( 7\left( 16.513878\right) +5\right) -\... ...}{\log 8} \\ && \\ &=&2.304684-0.901172=1.403512 \\ && \\ && \end{eqnarray*}


Right Side:         $\log _{8}\left( x+2\right) $

\begin{eqnarray*}&=&\log _{8}\left( 16.513878+2\right) \\ && \\ &=&\log _{8}\l... ... \left( 18.513878\right) }{\log 8} \\ && \\ &=&1.403512 \\ && \end{eqnarray*}


Since the left side of the original equation equals the right side of the original equation, the answer x=16.513878 is a solution.




Left Side: $\qquad \log _{8}\left( 7x+5\right) -\log _{8}\left( x-10\right) $

\begin{eqnarray*}&& \\ &=&\log _{8}\left( 7\left( -1.513878\right) +5\right) -\... ...6\right) -\log _{8}\left( -11.513878\right) \\ && \\ && \\ && \end{eqnarray*}

At this point in the check, we stop because we cannot take the logarithm of a negative number. The conclusion is that $ x\approx -1.513878$ is not a real solution.




Graphical Check:

You can also check your answer by graphing $\quad f(x)=\log _{8}\left( 7x+5\right) -\log _{8}\left( x-10\right) -\log _{8}\left( x+2\right) \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. You may have to modify the equation for your calculator first. Rewrite f(x) as

\begin{eqnarray*}&& \\ f(x) &=&\displaystyle \frac{\log \left( 7x+5\right) -\log \left( x-10\right) -\log \left( x+2\right) }{\log 8} \\ && \end{eqnarray*}


Note that the graph crosses the x-axis at one spot. This means that there is just one real solutions. The graph crosses the x-axis at x=16.513878, and this means that x=16.513878 is the real solution.

If you would like to review the solution to problem 8.5d, click on solution.


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