How do you solve ${log x}^{2} = {(log x)}^{2}$ $log x^2 = (log x)^2$ ?

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Answer 1 · 2015-12-01T21:59:46

$x = 100$ $x=100$

Rewrite as:

$2 log x = {(log x)}^{2}$ $2logx=(logx)^2$

$0 = {(log x)}^{2} - 2 log x$ $0=(logx)^2-2logx$

$0 = log x (log x - 2)$ $0=logx(logx-2)$

Set each portion equal to $0$ $0$ .

$\Rightarrow log x = 0$ $=>logx=0$

$x$ $x$ does not exist.

$\Rightarrow log x - 2 = 0$ $=>logx-2=0$

$log x = 2$ $logx=2$

${log}_{10} x = 2$ $log_10x=2$

$10^{{log}_{10} x} = 10^{2}$ $10^(log_10x)=10^2$

$x = 100$ $x=100$

How do you solve logx2=(logx)2log x^2 = (log x)^2?