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 8.6c:

$\log _{\displaystyle \frac{4}{3}}\left( 4x^{2}-7x-10\right) =20$

Answers: The exact answers are $x=\displaystyle \frac{7+\sqrt{209+16\cdot \left( \displaystyle \frac{4}{3}\right) ^{20}}}{8}.$ The approximate answers are $x\approx 9.935896\quad$ and $\quad -8,185896.$

Solution:

The above equation is valid only if $\quad \log _{\displaystyle \frac{4}{3}}\left( 4x^{2}-7x-10\right) =20$ is valid. The term $\log _{\displaystyle \frac{4}{3}}\left( 4x^{2}-7x-10\right)$ 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 $\displaystyle \frac{7-\sqrt{209}}{8}$ or greater than $\displaystyle \frac{7+\sqrt{209}}{8}.$

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

$\begin{eqnarray*}&& \\ \log _{\displaystyle \frac{4}{3}}\left( 4x^{2}-7x-10\rig... ...( 10+\left( \displaystyle \frac{4}{3}\right) ^{20}\right) \\ && \end{eqnarray*}$

$\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{7\pm \sqrt{49+4\left( 4\right) ... ...ac{4}{3}\right) ^{20}}}{8}\approx -8,185896 \\ && \\ && \\ && \end{eqnarray*}$

The exact answers are $x=\displaystyle \frac{7\pm \sqrt{209+16\cdot \left( \displaystyle \frac{4}{3} \right) ^{20}}}{8}.$ The approximate answers are $x\approx 9.935896$ and -8,185896.

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 $\quad x=\displaystyle \frac{7+\sqrt{209+16\cdot \left( \displaystyle \frac{4}{3 }\right) ^{20}}}{8}$ by substituting 9.935896 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: $\qquad \log _{\displaystyle \frac{4}{3}}\left( 4x^{2}-7x-10\right)$ $\begin{eqnarray*}&& \\ &=&\log _{\displaystyle \frac{4}{3}}\left( 4\left( 9.935... ...displaystyle \frac{4}{3}\right) } \\ &\approx &20 \\ && \\ && \end{eqnarray*}$

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 9.935896 for x, then x=9.935896 is a solution.

Check the answer $\quad x=\displaystyle \frac{7-\sqrt{209+16\cdot \left( \displaystyle \frac{4}{3 }\right) ^{20}}}{8}$ by substituting -8,185896 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: $\qquad \log _{\displaystyle \frac{4}{3}}\left( 4x^{2}-7x-10\right)$ $\begin{eqnarray*}&& \\ &=&\log _{\displaystyle \frac{4}{3}}\left( 4\left( -8,18... ...style \frac{4}{3}\right) } \\ && \\ &\approx &20 \\ && \\ && \end{eqnarray*}$

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 -8,185896 for x, then x=-8,185896 is a solution.

Graphical Check:

You can also check your answer by graphing

$\quad f(x)=\log _{\displaystyle \frac{4}{3} }\left( 4x^{2}-7x-10\right) -20\quad$ (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 9.935896 and -8,185896. This means that 9.935896and -8,185896 are the real solutions.

If you have trouble graphing the above problem, you might try graphing the equivalent function

$\begin{eqnarray*}&& \\ f(x) &=&\displaystyle \frac{\log \left( 4x^{2}-7x-10\rig... ...\log \left( \displaystyle \frac{4}{3} \right) }-20 \\ && \\ && \end{eqnarray*}$

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

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