`log_3(x^2) = 4.5` Solve the equation accurate to three decimal places

Saturday, December 20, 2014

To solve a logarithmic equation, we may simplify or rewrite it using the properties of logarithm.

For the given problem $\displaystyle{\log}_{{{3}}}{\left({{x}}^{{2}}\right)}={4.5}$ , we may apply the property:

$\displaystyle{{a}}^{{{\left({\log}_{{{a}}}{\left({x}\right)}\right)}}}={x}$

The "log" cancels out which we need to accomplish on the left side of the equation.

Raising both sides by the base of 3:

$\displaystyle{{3}}^{{{\left({\log}_{{3}}{\left({{x}}^{{2}}\right)}\right)}}}={{3}}^{{{4.5}}}$

$\displaystyle{{x}}^{{2}}={{3}}^{{{4.5}}}$

Taking the square root on both sides:

$\displaystyle\sqrt{{{{x}}^{{2}}}}=\pm\sqrt{{{{3}}^{{{4.5}}}}}$

$\displaystyle{x}=\pm{11.84466612}$

Rounded off to three decimal places:

$\displaystyle{x}=\pm{11.845}$ .

Plug-in the x-values to check if they are the real solution:

$\displaystyle{\log}_{{3}}{\left({{11.845}}^{{2}}\right)}={4.5}$ so x = 11.845 is a real solution.

Now let $\displaystyle{x}=-{11.845}$

$\displaystyle{\log}_{{3}}{\left({{\left(-{11.845}\right)}}^{{2}}\right)}$

$\displaystyle{\log}_{{3}}{\left({140.304025}\right)}={4.5}$ so x = -11.845 is a real solution.

So, x= 11.845, x = -11.845 are both solutions.

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