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 8.5a:

$\log _{6}\left( x-2\right) +\log _{6}\left( x+4\right) =\log _{6}\left( x-1\right) $

Answers: The exact answer is $x=\displaystyle \frac{-1+\sqrt{29}}{2}.$ The approximate answer is $x\approx 2.192582. $

Solution:

The above equation is valid only if each of the three terms is valid. The term $\log _{6}\left( x-2\right) $ is valid if x>2. The term $\log _{6}\left( x+4\right) $ is valid if x>-4. The term $\log _{6}\left( x-1\right) $ is valid if x>1. Therefore, the equation is valid when the domain is the set of real numbers is greater than2, greater than -4, and greater than1. This means that the equation is valid if we restrict the domain to the set of real numbers greater than2.

Simplify the equation and solve.

\begin{eqnarray*}&& \\ \log _{6}\left( x-2\right) +\log _{6}\left( x+4\right) &... ... \left( x-2\right) \left( x+4\right) &=&\left( x-1\right) \\ && \end{eqnarray*}
\begin{eqnarray*}&& \\ x^{2}+2x-8 &=&x-1 \\ && \\ && \\ x^{2}+x-7 &=&0 \\ &... ...rac{-1\pm \sqrt{1-4\left( 1\right) \left( -7\right) }}{2} \\ && \end{eqnarray*}
\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{-1\pm \sqrt{29}}{2} \\ && \\ ... ...yle \frac{-1-\sqrt{29}}{2}\approx -3.192582 \\ && \\ && \\ && \end{eqnarray*}

The exact answer is $x=\displaystyle \frac{-1+\sqrt{29}}{2}.$ The approximate answer is $x\approx 2.192582. $

These answer may or may not be the solution to the original equation. You must it in the original equation, either by numerical substitution or by graphing.

Numerical Check:

Left Side: $\qquad \log _{6}\left( x-2\right) +\log _{6}\left( x+4\right) $

\begin{eqnarray*}&& \\ &=&\log _{6}\left( 2.192582-2\right) +\log _{6}\left( 2.... ...92582\right) }{\log 6} \\ && \\ &=&-0.919338+1.017632=0.098294 \end{eqnarray*}


Right Side: $\qquad \log _{6}\left( x-1\right) =\log _{6}\left( 2.192582-1\right) =\log _{6}... ...2582\right) =\displaystyle \frac{\log \left( 1.192582\right) }{\log 6}=0.098295$

The left side does not equal the right side exactly because we rounded the numbers. However, it is close enough to verify our answers.Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 2.192582 for x, then x=2.192582is a solution.



Graphical Check:

You can also check your answer by graphing $\quad f(x)=\log _{6}\left( x-2\right) +\log _{6}\left( x+4\right) -\log _{6}\left( x-1\right) \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. You may have to modify the equation for your calculator first. Rewrite f(x) as

\begin{eqnarray*}&& \\ f(x) &=&\displaystyle \frac{\log \left( x-2\right) }{\lo... ...g 6}-\displaystyle \frac{\log \left( x-1\right) }{\log 6} \\ && \end{eqnarray*}


Note that the graph crosses the x-axis at 2.192582. This means that 2.192582 is the real solution.

If you would like to review the solution to problem 8.5b, click on solution.


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


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.

This site was built to accommodate the needs of students. The topics and problems are what students ask for. We ask students to help in the editing so that future viewers will access a cleaner site. If you feel that some of the material in this section is ambiguous or needs more clarification, or you find a mistake, please let us know by e-mail.



[Algebra] [Trigonometry]
[Geometry] [Differential Equations]
[Calculus] [Complex Variables] [Matrix Algebra]

S.O.S. MATHematics home page



Copyright © 1999-2004 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA