How do you solve $log_7x=log_2 9$ ?

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Answer 1 · 2016-03-25T08:52:36

$x=477.59$

As $log_b x=log_ax/log_ab$ , we have

$log_7 x=log_2 9$ is $log_10x/log_10 7=log_10 9/log_10 2$ i.e.

$logx/log7=log9/log2$

Hence $logx=(log9*log7)/log2=(0.9542*0.8451)/0.3010$ or

$logx=2.679$ or $x=10^(2.679)$ or

$x=477.59$

How do you solve log_7x=log_2 9log_7x=log_2 9?