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 function.



Solve for x in the following equation.


Problem 4:

$\ln \left( x-4\right) +\ln \left( 6x-5\right)
=\ln \left( 4x-3\right) +\ln \left( x-1\right) $



Answer:         The exact answer is $x=\displaystyle \frac{11+\sqrt{87}}{2};$ the approximate answer is $x\approx 10.1636895265.\bigskip\bigskip $

Solution:




The above equation is valid only if all of the terms are valid. The first term is valid if $x-4>0\longrightarrow $ x>4, the second term is valid if $
6x-5>0\longrightarrow x>\displaystyle \frac{5}{6},$ the third term is valid if $
4x-3>0\longrightarrow x>\displaystyle \frac{3}{4},$ and the fourth term is valid if $
x-1>0\longrightarrow x>1$. Therefore, the equation is valid when all four of these conditions are met, or when x>4. The domain is the set of real numbers greater than 4.



Simplify both sides of the equation using the rules of logarithms.

\begin{eqnarray*}&& \\
\ln \left( x-4\right) +\ln \left( 6x-5\right) &=&\ln \le...
...left( 4x-3\right) \left(
x-1\right) \right] \\
&& \\
&& \\
&&
\end{eqnarray*}
Recall that if $\ln (a)=\ln (b),$ then a = b. Therefore, if

\begin{eqnarray*}&& \\
\ln \left[ \left( x-4\right) \left( 6x-5 \right) \right]...
...( 6x-5\right) &=&\left( 4x-3\right) \left(
x-1\right) \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
6x^{2}-29x+20 &=&4x^{2}-7x+3 \\
&& \\
&& \\
2x^{2}-22x+17 &=&0 \\
&& \\
&& \\
&&
\end{eqnarray*}


Use the Quadratic Formula to solve for x.

\begin{eqnarray*}&& \\
If\qquad 2x^{2}-22x+17 &=&0 \\
&& \\
&& \\
x &=&\disp...
...left( 2\right) \left(
17\right) }}{2\left( 2\right) } \\
&& \\
\end{eqnarray*}



\begin{eqnarray*}&& \\
x &=&\displaystyle \frac{22\pm \sqrt{348}}{4} \\
&& \\
&& \\
x &=&\displaystyle \frac{22\pm 2\sqrt{87}}{4} \\
&& \\
\end{eqnarray*}

\begin{eqnarray*}&& \\
x &=&\displaystyle \frac{11\pm \sqrt{87}}{2} \\
&& \\
\end{eqnarray*}

\begin{eqnarray*}&& \\
x &=&\displaystyle \frac{11+\sqrt{87}}{2}\approx 10.163685265 \\
&& \\
\end{eqnarray*}

\begin{eqnarray*}&& \\
x &=&\displaystyle \frac{11-\sqrt{87}}{2}\approx 0.83631...
...\frac{11-\sqrt{87}}{2}\textbf{
is not a solution.} \\
&& \\
&&
\end{eqnarray*}


The exact answer is $x=\displaystyle \frac{11+\sqrt{87}}{2};$ the approximate answer is $
x\approx 10.163685265.\bigskip\bigskip\bigskip $

Check the answer x=10.163685265 by substituting 10.163685265 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.        


Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 10.163685265 for x, then x = 10.163685265 is a solution.





You can also check your answer by graphing $\quad f(x)=\ln \left( x-4\right)
+\ln \left( 6x-5\right) -\ln \left( 4x-3\right) -\ln \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. Note that the graph crosses the x-axis at 10.163685265. This means that 10.163685265 is the real solution.







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