How do you solve : log(x^2)=(log(x))^2 ? thanks

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Answer 1 · 2018-04-11T22:05:47

$x = 1, x = 100$ $color(blue)(x=1 , x=100)$

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

I am assuming these to be common logs ( base 10 ):

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

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

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

Factor:

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

$log (x) = 0$ $log(x)=0$

Using antilogarithm:

$10^{log (x)} = 10^{0}$ $10^(log(x))=10^0$

$x = 10^{0} = 1$ $x=10^0=1$

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

$log (x) = 2$ $log(x)=2$

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

$x = 10^{2} = 100$ $x=10^2=100$

$x = 1, x = 100$ $color(blue)(x=1 , x=100)$