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 1:        

ln $\left( x^{2}-6x-16\right) =5$.

The above equation is valid only if

$$\begin{eqnarray*}&& \\ \left( x^{2}-6x-16\right) &>&0\rightarrow \left( x-8\rig... ...\\ &&or \\ && \\ x-8 &<&0\ and\ x+2<0\rightarrow x<-2. \\ && \end{eqnarray*}$$

The domain is the set of real numbers less than -2 or greater than 8.

Convert the equation to an exponential equation with base e.

$$\begin{eqnarray*}&& \\ ln\left( x^{2}-6x-16\right) &=&5 \\ && \\ && \\ e^{5}... ...&& \\ && \\ && \\ 0=x^{2}-6x-\left( 16+e^{5}\right) && \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ x=\displaystyle \frac{6\pm \sqrt{36+\left( 4\right) \lef... ...&& \\ x=\displaystyle \frac{6\pm 2\sqrt{25+e^{5}}}{2} && \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ x=3\pm \sqrt{25+e^{5}} && \\ && \\ && \\ x=3+\sqrt{25... ...\\ x=3-\sqrt{25+e^{5}}\approx -10.16864 && \\ && \\ && \\ && \end{eqnarray*}$$

The exact answers are $\ x=3\pm \sqrt{25+e^{5}}$ and the approximate answers are 16.16864 and -10.16864.

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

Numerical Check:

Check the answer $x=3+\sqrt{25+e^{5}}$ by substituting 16.16864 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:$\qquad$ln $\left( x^{2}-6x-16\right) =$ln $\left( \left( 16.16864\right) ^{2}-6\left( 16.16864\right) -16\right) =4.9999995633\approx 5$

• Right Side:$\qquad 5$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 16.16864 for x, then x=16.16864 is a solution.

Check the answer $x=3-\sqrt{25+e^{5}}$ by substituting -10.16864 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:$\qquad$ln $\left( x^{2}-6x-16\right) =$ln $\left( \left( -10.16864\right) ^{2}-6\left( -10.16864\right) -16\right) =4.9999995633\approx 5$

• Right Side:$\qquad 5$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -10.16864 for x, then x=-10.16864 is a solution.

Graphical Check:

You can also check your answer by graphing

$\quad f(x)=ln\left( x^{2}-6x-16\right) -5\quad$(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 16.16864 and -10.16864. This means that 16.16864 and -10.16864 are the real solutions.

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