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 1:

$\log _{5}\left( x-2\right) +\log _{8}\left( x-4\right) =\log _{6}\left( x-1\right)$

The above equation is valid only if all the terms in the equation are valid. The term $\log _{5}\left( x-2\right)$ is valid if $\left( x-2\right) >0\longrightarrow x>2;$ the term $\log _{8}\left( x-4\right)$ is valid if $x-4>0\longrightarrow x>4;$ and the term $\log _{6}\left( x-1\right)$ is valid if $x-1>0\longrightarrow x>1..$ The domain is the set of real numbers that are greater than 2, greater than 4, and greater than 1. The domain is therefore the set of real numbers greater than 4.

This is a complex equation because all the bases are different. Let's solve it first by graphing; we do this by changing all the bases to either 10 or e. Why do we change the bases to either 10 or e? Because most calculators have these functions.

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

$$\begin{eqnarray*}&& \\ \log _{5}\left( x-2\right) +\log _{8}\left( x-4\right) &... ...tyle \frac{\ln \left( x-1\right) }{\ln (6)} \\ && \\ && \\ && \end{eqnarray*}$$

Rewrite the equation as $$\frac{\ln \left( x-2\right) }{\ln (5)}+\displaystyle \frac{\ln \l... ...x-4\right) }{\ln (8)}-\displaystyle \frac{\ln \left( x-1\right) }{\ln (6)} =0.$$

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

Then $f\left( x\right) =\displaystyle \frac{\ln \left( x-2\right) }{\ln (5)}+\display... ...eft( x-4\right) }{\ln (8)}-\displaystyle \frac{\ln \left( x-1\right) }{\ln (6)}$ 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 just 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=4. This is consistent with our finding that the domain of the original equation is the set of real numbers greater than 4.

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 about 5.1797. This means that the equation has one real solution at x=5.1797.

The algebraic solution is too difficult for a beginning student. Clearly, graphing is the easiest way to determine the answer.

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