How do you solve ${log}_{121} x = 2$ $log_(121)x= 2$ ?

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Answer 1 · 2016-02-01T19:44:57

$x = 14641$ $x=14641$

Let's use both sides of the equation as exponents with a base $121$ $121$ :
$121^{{log}_{121} (x)} = 121^{2}$ $121^(log_121(x)) = 121^2$

On the left, by definition of a logarithm we have just an unknown variable $x$ $x$ :
$x = 121^{2} = 14641$ $x = 121^2 = 14641$

How do you solve log121x=2log_(121)x= 2?