Question

Find the value of x if
$log_{ \frac{1}{3} }( {x}^{2} + 8) = - 2$
​

Answer 1

Answer :-

x = + 1 or x = - 1

Solution :-

$\displaystyle \implies log_{ \frac{1}{3} }( {x}^{2} + 8) = - 2$

Wrinting it in exponential form

$\displaystyle \implies \bigg( \frac{1}{3} \bigg)^{ - 2} = {x}^{2} + 8$

[ Because log_a N = x then a^x = N ]

$\displaystyle \implies (3)^{2} = {x}^{2} + 8$

[ Because 1 / a^( - n ) = a^n ]

$\displaystyle \implies 9 = {x}^{2} + 8$

$\displaystyle \implies 9 - 8= {x}^{2}$

$\displaystyle \implies 1= {x}^{2}$

$\displaystyle \implies {x}^{2} = 1$

$\displaystyle \implies x = \sqrt{1}$

$\displaystyle \implies x = \pm1$

Therefore the value of x is 1 or - 1.