How do you solve $Log(x+2)+log(x-1)=4$ ?

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Answer 1 · 2018-05-03T12:01:04

$x = \frac{-1 \pm sqrt(9 + 4 e^4)}{2}$

Use the rule that states

$log(a) + log(b) = log(ab)$

to get

$log((x+2)(x-1)) = 4$

exponential to both sides:

$(x+2)(x-1) = e^4$

Expand the parenthesis and bring everything to left side:

$x^2+x-2 - e^4 = 0$

Now this is a standard quadratic equation $ax^2+b+c=0$ , with $a=1$ , $b=1$ and $c=-2-e^4$ .

Pull these values into the formula

$x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

to get the answer.

How do you solve Log(x+2)+log(x-1)=4Log(x+2)+log(x-1)=4?