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 8.7a:

$5\log _{3}\left( 4x^{2}-7x-10\right) =20$

Answers: The exact answers are $x=\displaystyle \frac{7\pm \sqrt{1505}}{8}.$ The approximate answers are $x\approx 5.724291$ and -3.974291

Solution:

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

Isolate the log term.        

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

Convert the equation to an exponential equation with base 3.        

3⁴=4x²-7x-10

Set the equation equal to zero.        

4x²-7x-91.

Solve for x.        

$x=\displaystyle \frac{7\pm \sqrt{\left( -7\right) ^{2}-4\left( 4\right) \left( -91\right) }}{2\left( 4\right) }=\displaystyle \frac{7\pm \sqrt{1505}}{8}$

The exact answers are $x=\displaystyle \frac{7\pm \sqrt{1505}}{8}.$ The approximate answers are $x\approx 5.724291$ and -3.974291

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 $\quad$ $x=\displaystyle \frac{7+\sqrt{1505}}{8}$ by substituting $x\approx 5.724291$ 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: $\qquad 5\log _{3}\left( 4x^{2}-7x-10\right)$
  $$\begin{eqnarray*}&& \\ &=&5\log _{3}\left( 4\left( 5.724291\right) ^{2}-7\left(... ...left( 80.999993\right) }{\log \left( 3\right) } \\ && \\ &=&20 \end{eqnarray*}$$
  
  

• 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 5.724291 for x, then x=5.724291 is a solution.

Check the answer $\quad$ $x=\displaystyle \frac{7-\sqrt{1505}}{8}$ by substituting $x\approx -3.974291$ 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: $\qquad 5\log _{3}\left( 4x^{2}-7x-10\right)$
  $$\begin{eqnarray*}&& \\ &=&5\log _{3}\left( 4\left( -3.974291\right) ^{2}-7\left... ...left( 80.999993\right) }{\log \left( 3\right) } \\ && \\ &=&20 \end{eqnarray*}$$
  
  

• 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 -3.974291 for x, then x=-3.974291 is a solution.

Graphical Check:

You can also check your answer by graphing

$\quad f(x)=5\log _{3}\left( 4x^{2}-7x-10\right) -20\quad$ (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 5.724291 and -3.974291. This means that 5.724291 and -3.974291 are the real solutions. If you cannot graph the function on your calculator, you might try graphing the function

$f(x)=5\displaystyle \frac{\log \left( 4x^{2}-7x-10\right) }{\log (3)}-20$

If you would like to review the solution to problem 8.7b, click on solution.

If you would like to go back to the previous section, click on previous.

If you would like to go back to the equation table of contents, click on contents.

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