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 8.5b:

$\log _{3}\left( 7x-3\right) +\log _{3}\left(
x+15\right) =\log _{3}\left( x-5\right) $


Answers: There are no real solutions.


Solution:

The above equation is valid only if each of the three terms is valid. The term $\log _{3}\left( 7x-3\right) $ is valid if $7x-3>0\rightarrow x>\displaystyle \frac{3
}{7}.$ The term $\log \log _{3}\left( x+15\right) $ is valid if $
x+15>0\rightarrow x>-15.$ The term $\log \log _{3}\left( x-5\right) $ is valid if $x-5>0\rightarrow x>5.$ Therefore, the equation is valid when the domain is the set of real numbers is greater than $\displaystyle \frac{3}{7}$, greater than -15, and greater than 5. This means that the equation is valid if we restrict the domain to the set of real numbers greater than 5.


Simplify the equation and solve.

\begin{eqnarray*}&& \\
\log _{3}\left( 7x-3\right) +\log _{3}\left( x+15\right)...
...c{-101-\sqrt{11,321}}{14}\approx -14.814299 \\
&& \\
&& \\
&&
\end{eqnarray*}

Neither answer is greater than 5. Therefore, there are no real answers.



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


Numerical Check:

Left Side: $\qquad \log _{3}\left( 7x-3\right) +\log _{3}\left( x+15\right) $

\begin{eqnarray*}&& \\
&=&\log _{3}\left( 7\left( 0.385728\right) -3\right) +\l...
...
&=&\log _{3}\left( -0.299904\right) +\log _{3}15.385728 \\
&&
\end{eqnarray*}


As this point we stop our check because you cannot take the logarithm of a negative number. This means that x=0.385728 is not a solution.




Left Side: $\qquad \log _{3}\left( 7x-3\right) +\log _{3}\left( x+15\right) $

\begin{eqnarray*}&& \\
&=&\log _{3}\left( 7\left( -14.814299\right) -3\right) +...
...left( -103.700093\right) +\log _{3}\left( 0.185701\right) \\
&&
\end{eqnarray*}


As this point we stop our check because you cannot take the logarithm of a negative number. This means that x=-14.814299 is not a solution.



Graphical Check:

You can also check your answer by graphing $\quad f(x)=\log _{3}\left(
7x-3\right) +\log _{3}\left( x+15\right) -\log _{3}\left( x-5\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-3\right) +\log \left( x+15\right) -\log \left(
x-5\right) }{\log 3} \\
&&
\end{eqnarray*}

Note that the graph never crosses the x-axis. This means that there are no real solutions.


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


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