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# Solving Logarithmic Equations

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

## Solving a logarithmic equation Example

Solve the logarithmic equation:
log (3x+1) = 2
The base of common logarithms is 10.
3x + 1 = 102                     The equation is written in exponential form.
3x + 1 = 100
3x  = 99   $\rightarrow$      x = 33

## Properties of logarithms in solving a logarithmic equation

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

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 ( x2 - 4) = 0
x2 - 4  = 100                                                               Exponential form
x2 - 4  = 1
x2   = 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.