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 8.6b:

$\ln \left( 4x^{2}-7x-10\right) =20$

Answers:The exact answers are $x=\displaystyle \frac{7\pm \sqrt{ 209+16\cdot e^{20}}}{8}.$ The approximate answers are $x\approx 11,014.1079116\quad$and $\quad -11,012.3580457.$

Solution:

The above equation is valid only if $\quad$ln $\left( 4x^{2}-7x-10\right) =20$ is valid. The term ln $\left( 4x^{2}-7x-10\right)$ is valid if $\left( 4x^{2}-7x-10\right) >0\longrightarrow x>\displaystyle \frac{7+\sqrt{209}}{8}\approx 2.6821$ or $x<\displaystyle \frac{7-\sqrt{209}}{8}\approx -0.9321.$ Therefore, the equation is valid when the domain is the set of real numbers less than $$\frac{7-\sqrt{ 209}}{8}$$ or greater than $$\frac{7+\sqrt{209}}{8}.$$

Covert the logarithmic equation to an exponential equation with base e.

$$\begin{eqnarray*}&& \\ ln\left( 4x^{2}-7x-10\right) &=&20 \\ && \\ && \\ e^{... ...=&4x^{2}-7x-10 \\ && \\ && \\ 0 &=&4x^{2}-7x-10-e^{20} \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ 0 &=&4x^{2}-7x-\left( 10+10^{20}\right) \\ && \\ && \\... ...m \sqrt{49+4\left( 4\right) \left( 10+e^{20}\right) }}{8} \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{7\pm \sqrt{209+16\cdot e^{20}}}... ...+16\cdot e^{20}}}{8}\approx -11,012.3580457 \\ && \\ && \\ && \end{eqnarray*}$$

The exact answers are $x=\displaystyle \frac{7\pm \sqrt{ 209+16\cdot e^{20}}}{8}.$ The approximate answers are $x\approx 11,014.1079116$ and -11,012.3580457.

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 $\quad x=\displaystyle \frac{7+\sqrt{209+16\cdot e^{20}}}{8}$ by substituting 11,014.1079116 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: $\qquad \ln \left( 4x^{2}-7x-10\right)$ $$\begin{eqnarray*}&& \\ &=&\ln \left( 4\left( 11,014.1079116\right) ^{2}-7\left(... ....1079116\right) -10\right) \\ && \\ &\approx &20 \\ && \\ && \end{eqnarray*}$$

• Right Side:$\qquad 20$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 11,014.1079116 for x, then x=11,014.1079116 is a solution.

Check the answer $\quad x=\displaystyle \frac{7-\sqrt{209+16\cdot e^{20}}}{8}$ by substituting -11,012.3580457 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: $\qquad \log \left( 4x^{2}-7x-10\right)$ $$\begin{eqnarray*}&& \\ &=&\log \left( 4\left( -11,012.3580457\right) ^{2}-7\lef... ....3580457\right) -10\right) \\ && \\ &\approx &20 \\ && \\ && \end{eqnarray*}$$

• Right Side:$\qquad 20$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -11,012.3580457 for x, then x=-11,012.3580457 is a solution.

Graphical Check:

You can also check your answer by graphing

$\quad f(x)=\ln \left( 4x^{2}-7x-10\right) -20\quad$ (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 11,014.1079116 and -11,012.3580457. This means that 11,014.1079116 and -11,012.3580457 are the real solutions.

If you would like to review the solution to problem 8.6c, click on solution.

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