How do you solve $log x = log 3 + 2 log x$ $log x = log 3 + 2 log x$ ?

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How do you solve $log x = log 3 + 2 log x$ $log x = log 3 + 2 log x$ ?

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Answer 1 · 2016-04-23T15:57:56

$x = 3^{- 1} = \frac{1}{3}$ $x=3^-1=1/3$

Explanation:

$\Rightarrow 2 log x - log x = - log 3$ $=>2logx-logx =-log3$
$\Rightarrow log x = {log 3}^{- 1}$ $=>logx =log3^-1$
$x = 3^{- 1} = \frac{1}{3}$ $x=3^-1=1/3$

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How do you solve $log x = log 3 + 2 log x$ $log x = log 3 + 2 log x$ ?

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How do you solve $log x = log 3 + 2 log x$ $log x = log 3 + 2 log x$ ?