How do you solve $Log(x+2)+Log(x-1)=Log(88)$ ?

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Answer 1 · 2015-12-15T02:39:10

$x=9$

Use the product rule of logarithms: $log(a)+log(b)=log(ab)$

Thus, the expression can be written as

$log((x+2)(x-1)=log(88)$

Distribute.

$log(x^2+x-2)=log(88)$

Raise both sides as the power of $10$ .

$10^(log(x^2+x-2))=10^(log(88))$

$x^2+x-2=88$

$x^2+x-90=0$

$(x+10)(x-9)=0$

$x+10=0$
or
$x-9=0$

$x=-10$
or
$x=9$

Plug in both potential values.

Notice that if you plug in $-10$ , you'd have to take the logarithm of a negative number, which is impossible.

Thus, the $-10$ answer is thrown out and all that's left is

$x=9$

How do you solve Log(x+2)+Log(x-1)=Log(88)Log(x+2)+Log(x-1)=Log(88)?