How do you solve $(log x)^2+ 3 log x - 10 = 0$ ?

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Answer 1 · 2016-03-02T15:29:52

$x=100$ and $x=10^-5$

From the given:

$(log x)^2+3*log x -10=0$

this is a quadratic equation in log x

Let y=log x

then

$y^2+3y-10=0$
solve by factoring

$(y-2)(y+5)=0$

we have $y=2$ and $y=-5$

Therefore

$log x=2$

$10^2=x$
$x=100$
~~~~~~~~~~~~~~
Also $y=-5$

and $log x =-5$

$10^-5=x$

$x=1/100000=0.00001$

$x=0.00001$

God bless...I hope the explanation is useful

How do you solve (log x)^2+ 3 log x - 10 = 0(log x)^2+ 3 log x - 10 = 0?