How do you solve $log 5 = 2 - log x$ $Log 5 = 2 - Log x$ ?

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Answer 1 · 2016-02-27T23:13:14

$x = \frac{e^{2}}{5}$ $x=e^2/5$

$log 5 = 2 - log x$ $log5=2−logx$

$log x = 2 - log 5$ $logx=2−log5$

$log x = log (e^{2}) - log 5$ $logx= log (e^2)-log5$

$log x = log (\frac{e^{2}}{5})$ $logx= log (e^2/5)$

$x = \frac{e^{2}}{5}$ $x=e^2/5$

How do you solve log5=2−logx Log 5 = 2 - Log x?