How do you solve $log_2x=log_5 3$ ?

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Answer 1 · 2016-05-26T10:19:53

$x=1.605$

$log_2x=log_53$ can be simplified using $log_ba=loga/logb$ . Hence it is

$logx/log2=log3/log5$

or $logx=log3/log5xxlog2$

or $logx=0.4771/0.6990xx0.3010$

Hence $x=10^(0.4771/0.6990xx0.3010)=1.605$

How do you solve log_2x=log_5 3log_2x=log_5 3?