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 4:

$\log _{102}\left( x^{2}-5x-11\right) +5=105$

The above equation is valid only if $\left( x^{2}-5x-11\right) >0$ or x<-4or x>9. The domain is the set of real numbers $\left( -\infty ,4\right) \cup \left( 9,\infty \right) .$

Simplify the left side of the equation using the rules of logarithms.

$$\begin{eqnarray*}&& \\ \log _{102}\left( x^{2}-5x-11\right) +5 &=&105 \\ && \\ && \\ \log _{102}\left( x^{2}-5x-11\right) &=&100 \\ && \\ \end{eqnarray*}$$

Convert the equation to an exponential equation with base 102. $$\begin{eqnarray*}&& \\ \log _{102}\left( x^{2}-5x-11\right) &=&100 \\ && \\ && \\ \left( 102\right) ^{100} &=&x^{2}-5x-11 \\ && \\ \end{eqnarray*}$$
$$\begin{eqnarray*}&& \\ x^{2}-5x-11-\left( 102\right) ^{100} &=&0 \\ && \\ && ... ...{2}-5x-\left( 11+\left( 102\right) ^{100}\right) &=&0 \\ && \\ \end{eqnarray*}$$
$$\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{5\pm \sqrt{\left( -5\right) ^{2... ...t{25+4\left( 11+\left( 102\right) ^{100}\right) }}{2} \\ && \\ \end{eqnarray*}$$
$$\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{5\pm \sqrt{2.8978584473\times 1... ...mes 10^{201}}}{2}\approx 2.69158802907\times 10^{100} \\ && \\ \end{eqnarray*}$$
$$\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{5-\sqrt{2.8978584473\times 10^{201}}}{2}\approx -2.69158802907\times 10^{100} \\ &&\\ \end{eqnarray*}$$

The exact answer is $x=\displaystyle \frac{5\pm \sqrt{2.8978584473\times 10^{201}}}{2},$the approximate answers are $x\approx \pm 2.69158802907\times 10^{100}.\bigskip\bigskip\bigskip$

These answers may or may not be the solutions to the original equation. You must check them in the original equation, either by numerical substitution or by graphing.

Numerical Check:

Check the answer $x=\displaystyle \frac{5+\sqrt{2.8978584473\times 10^{201}}}{2}$by substituting $2.69158802907\times 10^{100}$ in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.

• Left Side: $\qquad \log _{102}\left( x^{2}-5x-11\right) +5$
  $$\begin{eqnarray*}&& \\ &=&\log _{102}\left( \left( 2.69158802907\times 10^{100}... ...ft( 2.69158802907\times 10^{100}\right) -11\right) +5 \\ && \\ \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &\approx &\displaystyle \frac{\ln \left( 7.24464611823\t... ...{\ln \left( 102\right) }+5 \\ && \\ &\approx &100+5=105 \\ && \end{eqnarray*}$$

  
  

  
  

• Right Side: $\qquad 105$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value $2.69158802907\times 10^{100}$ for x, then $x=2.69158802907\times 10^{100}$ is a solution.

Check the answer $x=\displaystyle \frac{5-\sqrt{2.8978584473\times 10^{201}}}{2}$by substituting $-2.69158802907\times 10^{100}$ in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.

• Left Side: $\qquad \log _{102}\left( x^{2}-5x-11\right) +5$
  
  $$\begin{eqnarray*}&& \\ &=&\log _{102}\left( \left( -2.69158802907\times 10^{100... ...t( -2.69158802907\times 10^{100}\right) -11\right) +5 \\ && \\ \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &\approx &\displaystyle \frac{\ln \left( 7.24464611823\t... ...{\ln \left( 102\right) }+5 \\ && \\ &\approx &100+5=105 \\ && \end{eqnarray*}$$

  
  

  
  

• Right Side: $\qquad 105$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value $-2.69158802907\times 10^{100}$ for x, then $x=-2.69158802907\times 10^{100}$ is a solution.

Graphical Check:

You can also check your answer by graphing $\quad f(x)=\log _{102}\left( x^{2}-5x-11\right) -100\quad$(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 $\pm 2.69158802907\times 10^{100}$. This means that $\pm 2.69158802907\times 10^{100}$ are the real solutions.

You may have to change the original equation somewhat to graph it because most graphing calculators only have the natural log function and the common log function. Rewrite the original equation $f(x)=\log _{102}\left( x^{2}-5x-11\right) -100$ in the equivalent form $f(x)=\displaystyle \frac{\ln (x^{2}-5x-11)}{\ln \left( 102\right) }-100$ and graph it

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