How do you solve $(ln(x))^2 + ln(x) - 6 = 0$ ?

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Answer 1 · 2018-06-05T15:35:03

$x=e^(-3)=1/(e^3) or x=e^2$

We know that,

$(1)log_a y=X <=>y=a^X, AAa in RR^+ -{1},yinRR^+,X inRR$

$(2)color(red)(ln(y)=X<=>y=e^X ),where,yinRR^+ andcolor(red)( X inRR$

Here,

$(ln(x))^2+ln(x)-6=0$

Let,

$ln(x)=m$

So,

$m^2+m-6=0$

$m^2+3m-2m-6=0$

$m(m+3)-2(m+3)=0$

$(m+3)(m-2)=0$

$m+3=0 or m-2=0$

$m=-3 or m=2$

Subst. back, $m=ln(x)$

$:.lnx=-3 or lnx=2,where,color(red)( (-3) and 2inRR$

$=>x=e^(-3) or x=e^2$

How do you solve (ln(x))^2 + ln(x) - 6 = 0(ln(x))^2 + ln(x) - 6 = 0?