Logarithmic functions are inverse functions of exponential equations.An equation b = a^{x} can be written as a logarithmic equation as log_{a} b = x. As logarithms are used to solve an exponential equations, exponents are used to solve a logarithmic equation.

Solve the logarithmic equation:

log (3x+1) = 2

The base of common logarithms is 10.

3x + 1 = 10

^{2} The equation is written in exponential form.

3x + 1 = 100

3x = 99 $\rightarrow $ x = 33

## Properties of logarithms in solving a logarithmic equation

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At times it is required to simplify the given logarithmic equation to a simpler one, so that it can be translated into exponential form. The given equation is manipulated using properties of logarithm.

**Example:**log (7x+1) = log (x-2) +1

log (7x+1) - log (x-2) = 1 The logarithmic terms are grouped

$log\frac{7x+1}{x-2} =1$ Quotient rule for logarithms

$\frac{7x+1}{x-2} =10^{1}$ The equation is written in exponent form

$7x+1 = 10(x-2)$

$7x+1 = 10x -20$

$-3x = -21$ $\rightarrow $ $x = 7$

## Extraneous solutions for a logarithmic equation

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The domain of exponential functions is the set of all real numbers.
This against the logarithms are defined only for positive values. Hence
the solutions for logarithmic equations are to be verified for their
validity.

**Example:**log (x+2) + log(x-2) = 0

log [(x+2)(x-2)] = 0 Product rule

log ( x

^{2} - 4) = 0

x

^{2} - 4 = 10

^{0} Exponential form

x

^{2} - 4 = 1

x

^{2} = 5 $\rightarrow $ $x = \pm 5$

But the value x =-5 will render log(x-2) undefined. Hence x =-5 is an extraneous solutions and needs to be discarded.

^{} ^{}^{} The solutions of a logarithmic equation can be estimated using graphical method.

Consider the equation log (5-2x) = 0

To solve the equation graphically the x intercept of the graph y = 5-2x is to be found.The graph cuts the x axis at x =2.

The algebraical method also leads to the same solution.