How do you solve ${log}_{9} x = 2$ $log_9x=2$ ?

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Answer 1 · 2016-11-13T17:11:47

${log}_{9} x = 2$ $log_9 x = 2$

Use the log rule: If ${log}_{b} x = a$ $log_b x= a$ , then $b^{a} = x$ $b^a=x$

$9^{2} = x$ $9^2=x$

$x = 81$ $x=81$

Answer 2 · 2016-11-13T17:12:36

$x = 81$ $x=81$

From definition of logarithm

if $a^{m} = b$ $a^m=b$ then ${log}_{a} b = m$ $log_a b=m$

and vice versa i.e. if ${log}_{a} b = m$ $log_a b=m$ then $a^{m} = b$ $a^m=b$

Hence ${log}_{9} x = 2$ $log_9 x=2$ implies $9^{2} = x$ $9^2=x$

i.e. $x = 81$ $x=81$

How do you solve log9x=2log_9x=2?