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.