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 8.4b:

$\log _{12}\left( 4x-5\right) ^{2}=16$

Answer:

x=107,495,425.25 and $107,495,422.75\bigskip \bigskip$

Solution:

The above equation is valid only if the term $\quad \log _{12}\left( 4x-5\right) ^{2}$ is valid. The term $\log _{12}\left( 4x-5\right) ^{2}$ is valid if $\left( 4x-5\right) ^{2}>0\longrightarrow x\neq \displaystyle \frac{5}{4}.$Therefore, the equation is valid when $x\neq \displaystyle \frac{5}{4}.$ Another way of saying this is that the domain is the set of real numbers where $x\neq \displaystyle \frac{5}{4}.$

Convert the logarithmic equation to an exponential equation with base 12.

$$\begin{eqnarray*}&& \\ \log _{12}\left( 4x-5\right) ^{2} &=&16 \\ && \\ && \\ 12^{16} &=&\left( 4x-5\right) ^{2} \\ && \\ \end{eqnarray*}$$
$$\begin{eqnarray*}&& \\ \pm 12^{8} &=&4x-5 \\ && \\ && \\ 5\pm 12^{8} &=&4x \\ && \\ \end{eqnarray*}$$
$$\begin{eqnarray*}&& \\ \displaystyle \frac{5\pm 12^{8}}{4} &=&\displaystyle \fr... ...& \\ && \\ x &=&\displaystyle \frac{5\pm 12^{8}}{4} \\ && \\ \end{eqnarray*}$$
$$\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{5+12^{8}}{4}=107,495,425.25 \\ ... ...\ x &=&\displaystyle \frac{5-12^{8}}{4}=-107,495,422.75 \\ && \end{eqnarray*}$$

The exact answers are x=107,495,425.25 and x=-107,495,422.75. These answers may or may not be 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 x=107,495,425.25 by substituting 107,495,425.25in 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: $\qquad \log _{12}\left( 4x-5\right) ^{2}=\log _{12}\left( 4\left( 107,495,425.25\right) -5\right) ^{2}$

$$\begin{eqnarray*}&& \\ &=&\log _{12}\left( 4\left( 107,495,425.25\right) -5\rig... ... 10^{17}\right) }{\ln \left( 12\right) } \\ && \\ &=&16 \\ && \end{eqnarray*}$$

Right Side:$\qquad 16$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 107,495,425.25 for x, then x=107,495,425.25 is a solution.

• Check the answer x=-107,495,422.75 by substituting -107,495,422.75in 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: $\qquad \log _{12}\left( 4x-5\right) ^{2}=\log _{12}\left( 4\left( -107,495,422.75\right) -5\right) ^{2}$

$$\begin{eqnarray*}&& \\ &=&\log _{12}\left( 4\left( -107,495,422.75\right) -5\ri... ... 10^{17}\right) }{\ln \left( 12\right) } \\ && \\ &=&16 \\ && \end{eqnarray*}$$

Right Side:$\qquad 16$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 1107,495,422.75 for x, then x=-107,495,422.75 is a solution.

Note: If you had simplified the problem to $\log _{12}\left( 4x-5\right) =8,$you would have lost one of the answers.

Graphical Check:

You can also check your answer by graphing $\quad f(x)=\log _{12}\left( 4x-5\right) ^{2}-16\quad$(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 107,495,425.25 and -107,495,422.75. This means that 107,495,425.25 and -107,495,422.75 are the real solutions.

If you have trouble graphing the function $f(x)=\log _{12}\left( 4x-5\right) ^{2}-16\quad$, graph the equivalent function $f(x)=\displaystyle \frac{\ln 4x-5^{2}}{\ln \left( 12\right) }-16\quad$

If you would like to rewiew the solution to problem 8.4c, click on solution.

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