How do you solve ${(log x)}^{2} + log (x^{3}) + 2 = 0$ $(Logx)^2 + Log(x^3) + 2 = 0$ ?

Question

13318 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-14T00:11:56

${(log x)}^{2} + log (x^{3}) + 2 = 0$ $(log x)^2 +log(x^3)+2=0$

log(a^b)=blog(a)

${(log x)}^{2} + 3 log (x) + 2 = 0$ $(log x)^2 +3log(x)+2=0$

This is one expresssion of the type ay^2 +by +c=0

Now we solve the second degree equation:

$log (x) = \frac{- 3 \pm \sqrt{3^{2} - 4 \cdot 2}}{2}$ $log(x)=(-3+-sqrt(3^2-4*2))/2$

$log (x) = \frac{- 3 \pm 1}{2}$ $log(x)=(-3+-1)/2$

$log (x) = - \frac{2}{2} or log (x) = - \frac{4}{2}$ $log(x)=-2/2 or log(x)=-4/2$

$log (x) = - 1 or log (x) = - 2$ $log(x)=-1 or log(x)=-2$

$x = e^{- 1} or x = e^{- 2}$ $x=e^-1 or x=e^-2$

How do you solve (Logx)^2 + Log(x^3) + 2 = 0(logx)2+log(x3)+2=0(Logx)^2 + Log(x^3) + 2 = 0?