Solve the equation $(logx)^2-4logx=0$ ?

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Answer 1 · 2017-07-06T16:18:25

See below.

$(logx)^2-4logx=0$ or

$(logx-4)logx=0$ so we have the solutions

${(logx-4=0 -> x= e^4),(logx=0->x=1):}$

Answer 2 · 2017-07-06T16:22:01

$x=1$ or $x=e^4$

We have:

$(lnx)^2 = ln(x^4)$

Using the rules of logarithms, this can be written as:

$(lnx)^2 = 4lnx$

$:. (lnx)^2 - 4lnx = 0$
$:. lnx(lnx-4) = 0$

This would lead to two possible solutions:

Either:

$ln x = 0 => x=e^0$
$" " => x=1$

Or:

$lnx-4 =0 => ln x =4$
$" " => x=e^4$

Solve the equation (logx)^2-4logx=0(logx)^2-4logx=0?