How do you solve ${log}_{9} x^{2} = 5$ $\log _ { 9} x ^ { 2} = 5$ ?

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Answer 1 · 2017-12-07T03:43:16

$x = 243 or x = - 243$ $x=243 or x=-243$

${log}_{9} x^{2} = 5$ $log_9x^2=5$

${log}_{9} x^{2} = \frac{log (x^{2})}{log 9}$ $log_9x^2=log(x^2)/log9$

$\frac{log (x^{2})}{log 9} = 5$ $log(x^2)/log9=5$

Multiply both sides by log9

$log (x^{2}) = 5 log 9$ $log(x^2)=5log9$

$5 log 9 = log (9^{5}) = log (59049)$ $5log9=log(9^5)=log(59049)$

${log x}^{2} = log (59049)$ $logx^2=log(59049)$

Cancel logarithms by taking exp of both sides

$x^{2} = 59049$ $x^2=59049$

take square root of both sides

$x = 243 or x = - 243$ $x=243 or x=-243$

Answer 2 · 2017-12-07T08:48:52

$x = \pm 3^{5} = \pm 243$ $x=+-3^5=+-243$

we can use the definition of logs

${log}_{a} b = c \Rightarrow a^{c} = b$ $log_ab=c=>a^c=b$

${log}_{9} x^{2} = 5 \Rightarrow 9^{5} = x^{2}$ $log_9x^2=5=>9^5=x^2$

$x^{2} = 9^{5} = {(3^{2})}^{5}$ $x^2=9^5=(3^2)^5$

$\Rightarrow x = = \pm \sqrt{{(3^{2})}^{5}} = \pm {(\sqrt{3^{2}})}^{5}$ $=>x==+-sqrt((3^2)^5)=+-(sqrt(3^2))^5$

$:.x=+-3^5=+-243$

How do you solve \log _ { 9} x ^ { 2} = 5log9x2=5\log _ { 9} x ^ { 2} = 5?