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.

Example 3:

$\log _{8}\left( 7x+5\right) -\log _{8}\left( x-10\right) =\log _{8}\left( x+2\right)$

The above equation is valid only if all the terms in the equation are valid. The term $\log _{8}\left( 7x+5\right)$ is valid if $\left( 7x+5\right) >0\longrightarrow x>-\displaystyle \frac{5}{7};$ the term $\log _{8}\left( x+5\right)$is valid if $x-10>0\longrightarrow x>10;$ and the term $\log _{8}\left( x\right)$ is valid if $x+2>0\rightarrow x>-2.$ The domain is the set of real numbers that are greater than $-\displaystyle \frac{5}{7}$, greater than 10, and greater than -2. The domain is therefore the set of real numbers greater than 10.

Algebraic Method:

$$\begin{eqnarray*}&& \\ \log _{8}\left( 7x+5\right) -\log _{8}\left( x-10\right)... ...( 7x+5\right) }{\left( x-10\right) } &=&\left( x+2\right) \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ \left( x-10\right) \displaystyle \frac{\left( 7x+5\right... ...) \left( x+2\right) \\ && \\ && \\ 7x+5 &=&x^{2}-8x-20 \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ 0 &=&x^{2}-15x-25 \\ && \\ && \\ x &=&\displaystyle \frac{15\pm \sqrt{325}}{2} \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{15+\sqrt{325}}{2}\approx 16.513... ...15-\sqrt{325}}{2}\approx -1.51387818866> 10 \\ && \\ && \\ && \end{eqnarray*}$$

Although there are two solutions, only one of the solutions is in the domain of the problem. The exact solution to the problem is $x=\displaystyle \frac{15+\sqrt{325} }{2}$, and the approximate solution is $x\approx 16.5138781887.$

Graphing Method:

Change the original equation into an equation where all the logarithmic terms have base 10.

$$\begin{eqnarray*}&& \\ \log _{6}\left( 2x-3\right) +\log _{6}\left( x+5\right) ... ...og \left( x\right) }{\log \left( 3\right) } \\ && \\ && \\ && \end{eqnarray*}$$

Rewrite the equation as $$\frac{\log \left( 2x-3\right) }{\log \left( 6\right) }+\displayst... ...right) }-\displaystyle \frac{ \log \left( x\right) }{\log \left( 3\right) }=0.$$

Let's call the left side of the equation f(x) and the right side of the equation g(x).

Then $f\left( x\right) =\displaystyle \frac{\log \left( 2x-3\right) }{\log \left( 6\r... ...( 6\right) }-\displaystyle \frac{ \log \left( x\right) }{\log \left( 3\right) }$ and $g\left( x\right) =0.$Graph f(x) and g(x). We are looking for the point(s) of intersection, $\left( x,y\right) .$ The solution, if any, will be the value of x in the point(s) of intersection.

The graph of the right side of the equation is the set of points where the value of y equals zero. This is easy; it is the x-axis. We then look to see where the graph of f(x) crosses the x-axis.

Note that the graph only appears to the right of $x=\displaystyle \frac{3}{2}.$ This is consistent with our finding that the domain of the original equation is the set of real numbers greater than $$\frac{3}{2}$$.

The solution(s) to the original equation is the set of real numbers where f(x) crosses the x-axis (the x-intercepts are the solutions to the problem.) . You will note from the graph that f(x) crosses the x-axis at x=1.6735161761. This means that the equation has one real solution at $x\approx 1.6735161761$ .

Algebraic Method:

Change the bases of the logarithmic terms to 10.

$$\begin{eqnarray*}&& \\ \log _{6}\left( 2x-3\right) +\log _{6}\left( x+5\right) ... ...og \left( x\right) }{\log \left( 3\right) } \\ && \\ && \\ && \end{eqnarray*}$$

Multiply both sides of the equation by the factor $\left( \log \left( 6\right) \right) \left( \log \left( 3\right) \right) .$

$$\begin{eqnarray*}&& \\ && \\ \left( \log \left( 6\right) \right) \left( \log \... ...tyle \frac{\log \left( x\right) }{\log \left( 3\right) } \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ \log \left( 3\right) \cdot \log \left( 2x-3\right) +\log... ...right) } &=&\log \left( x\right) ^{\log \left( 6\right) } \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ \log \left( 2x-3\right) ^{\log \left( 3\right) }\cdot \l... ...&=&\left( x\right) ^{\log \left( 6\right) } \\ && \\ && \\ && \end{eqnarray*}$$

An algebraic solution, other than interpolation, is too difficult for a beginning student.

Interpolation Method:

Since the domain is the set of real numbers greater than $$\frac{3}{2},$$choose numbers larger than $$\frac{3}{2}.$$

• Let's start with x=1.6and evaluate f(1.6). Since we are looking for the value such that $f\left( value\right) =0,$ our target is 0.

  
  

  $$\begin{eqnarray*}f\left( x\right) &=&\displaystyle \frac{\log \left( 2x-3\right)... ...yle \frac{\log \left( 1.6\right) }{\log \left( 3\right) } \\ && \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &=&-0.898244401704+1.05319362417-0.427815739996 \\ && \\ && \\ &=&-0.27286651753 \\ && \\ && \\ &<&0 \end{eqnarray*}$$

  
  

  
  

• Let's now try x=10 and evaluate f(10). Since we are looking for the value such that $f\left( value\right) =0,$ our target is 0.

  
  

  $$\begin{eqnarray*}f\left( x\right) &=&\displaystyle \frac{\log \left( 2x-3\right)... ...tyle \frac{\log \left( 10\right) }{\log \left( 3\right) } \\ && \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &=&1.5812464746+1.51139159447-2.09590327429 \\ && \\ && \\ &=&0.99673479907 \\ && \\ && \\ &>&0 \\ && \\ && \\ && \end{eqnarray*}$$

  Since f(1.6)<0 and f(10)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=10. This means that we know at least one solution has a value symbol> than 1.6 and less than 10. Since $f\left( 1.6\right) =-0.27286651753$ is closer to zero than $f\left( 10\right) =0.99673479907$, we know that the answer most likely lies to the left of the midpoint of 1.6 and 10, or to the left of 7.4.

  
  

  
  

• Let's try x=4.5 (4.5 is located to the left of 7.4) and evaluate f(4.5). We are looking for a value of x such that $f\left( value\right) =0,$ our target is 0.

  
  

  $$\begin{eqnarray*}f\left( x\right) &=&\displaystyle \frac{\log \left( 2x-3\right)... ...yle \frac{\log \left( 4.5\right) }{\log \left( 3\right) } \\ && \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &=&1+1.25646987627-1.36907024643 \\ && \\ && \\ &=&0.887399629842 \\ && \\ && \\ &>&0 \\ && \\ && \\ && \end{eqnarray*}$$

  Since f(1.6)<0 and f(4.5)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=4.5. This means that we know at least one solution has a value greater than 1.6 and less than 4.5. Since $f\left( 1.6\right) =-0.27286651753$ is closer to zero than $f\left( 4.5\right) =0.887399629842$, we know that the answer most likely lies to the left of the midpoint of 1.6 and 4.5 or to the left of 3.05..

  
  

  
  

• Let's try x=2 and evaluate f(2). Since we are looking for the value such that $f\left( value\right) =0,$ our target is 0.

  
  

  $$\begin{eqnarray*}f\left( x\right) &=&\displaystyle \frac{\log \left( 2x-3\right)... ...tyle \frac{ \log \left( 2\right) }{\log \left( 3\right) } \\ && \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &=&0+1.0860331325-0.630929753571 \\ && \\ && \\ &=&0.455103378929 \\ && \\ && \\ &>&0 \\ && \\ && \\ && \end{eqnarray*}$$

  Since f(1.6)<0 and f(2)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=2. This means that we know at least one solution has a value greater than 1.6 and less than 2. Since $f\left( 1.6\right) =-0.27286651753$ is closer to zero than $f\left( 2\right) =0.455103378929$, we know that the answer most likely lies to the left of the midpoint of 1.6 and 2 or to the left of 1.8.

  
  

  
  

  
  

• Let's try x=1.7 and evaluate f(1.7). Since we are looking for the value such that $f\left( value\right) =0,$ our target is 0.

  
  

  $$\begin{eqnarray*}f\left( x\right) &=&\displaystyle \frac{\log \left( 2x-3\right)... ...yle \frac{\log \left( 1.7\right) }{\log \left( 3\right) } \\ && \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &=&-0.511391594469+1.06158642333-0.482998648873 \\ && \... ...\ &=&0.067196179988 \\ && \\ && \\ &>&0 \\ && \\ && \\ && \end{eqnarray*}$$

  Since f(1.6)<0 and f(1.7)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=1.7. This means that we know at least one solution has a value greater than 1.6 and less than 1.7. Since $=f\left( 1.7\right) =0.067196179988$ is closer to zero than $f\left( 1.6\right) =-0.27286651753,$ the answer is closer to 1.7 than to 1.6.

  
  

  
  

• Let's try x=1.68 and evaluate f(1.68). Since we are looking for the value such that $f\left( value\right) =0,$ our target is 0.

  
  

  $$\begin{eqnarray*}f\left( x\right) &=&\displaystyle \frac{\log \left( 2x-3\right)... ...le \frac{\log \left( 1.68\right) }{\log \left( 3\right) } \\ && \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &=&-0.570194417877+1.05991792993-0.472226461297 \\ && \... ...\ &=&0.017497050756 \\ && \\ && \\ &>&0 \\ && \\ && \\ && \end{eqnarray*}$$

  Since f(1.6)<0 and f(1.68)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=1.68. This means that we know at least one solution has a value greater than 1.6 and less than 1.68. Since $=f\left( 1.68\right) =0.017497050756$ is closer to zero than $f\left( 1.6\right) =-0.27286651753,$ the answer is closer to 1.68 than to 1.6.

  
  

  
  

• Let's try x=1.67 and evaluate f(1.67). Since we are looking for the value such that $f\left( value\right) =0,$ our target is 0.

  
  

  $$\begin{eqnarray*}f\left( x\right) &=&\displaystyle \frac{\log \left( 2x-3\right)... ...le \frac{\log \left( 1.67\right) }{\log \left( 3\right) } \\ && \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &=&-0.602095136038+1.05908180898-0.466792181116 \\ && \... ... &=&-0.009805508174 \\ && \\ && \\ &<&0 \\ && \\ && \\ && \end{eqnarray*}$$
  
  

  Since f(1.67)<0 and f(1.68)>0, we know that the graph of f(x) must cross the x-axis between x=1.67 and x=1.68. This means that we know at least one solution has a value greater than 1.67 and less than 1.68.

  
  

  
  

• You can now try x=1.675 and place the solution between x=1.67 and x=1.675 or between x=1.675 and x=1.68. You con continue in this fashion until your approximation of the answer is carried to as many decimals as your problems requires. Let's try x=1.68= and evaluate f(1.68). Since we are looking for the value such that $f\left( value\right) =0,$ our target is 0.

  
  

  $$\begin{eqnarray*}f\left( x\right) &=&\displaystyle \frac{\log \left( 2x-3\right)... ...le \frac{\log \left( 1.68\right) }{\log \left( 3\right) } \\ && \end{eqnarray*}$$
  $$\begin{eqnarray*}&& \\ &=&-0.570194417877+1.05991792993-0.472226461297 \\ && \... ...\ &=&0.017497050756 \\ && \\ && \\ &>&0 \\ && \\ && \\ && \end{eqnarray*}$$

  Since f(1.6)<0 and f(1.68)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=1.68. This means that we know at least one solution has a value greater than 1.6 and less than 1.68. Since $=f\left( 1.68\right) =0.017497050756$ is closer to zero than $f\left( 1.6\right) =-0.27286651753,$ the answer is closer to 1.68 than to 1.6.

If you would like to test yourself by working some problems similar to this example, click on problem.

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