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

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Answer 1 · 2018-05-03T12:01:04

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

Use the rule that states

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

to get

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

exponential to both sides:

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

Expand the parenthesis and bring everything to left side:

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

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

Pull these values into the formula

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

to get the answer.

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