How do you solve ${log}_{2} x = {log}_{5} 3$ $log_2x=log_5 3$ ?

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Answer 1 · 2016-05-26T10:19:53

$x = 1.605$ $x=1.605$

${log}_{2} x = {log}_{53}$ $log_2x=log_53$ can be simplified using ${log}_{b} a = \frac{log a}{log b}$ $log_ba=loga/logb$ . Hence it is

$\frac{log x}{log 2} = \frac{log 3}{log 5}$ $logx/log2=log3/log5$

or $log x = \frac{log 3}{log 5} \times log 2$ $logx=log3/log5xxlog2$

or $log x = \frac{0.4771}{0.6990} \times 0.3010$ $logx=0.4771/0.6990xx0.3010$

Hence $x = 10^{\frac{0.4771}{0.6990} \times 0.3010} = 1.605$ $x=10^(0.4771/0.6990xx0.3010)=1.605$

How do you solve log2x=log53log_2x=log_5 3?