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

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Answer 1 · 2015-12-15T02:39:10

$x = 9$ $x=9$

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

Thus, the expression can be written as

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

Distribute.

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

Raise both sides as the power of $10$ $10$ .

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

$x^{2} + x - 2 = 88$ $x^2+x-2=88$

$x^{2} + x - 90 = 0$ $x^2+x-90=0$

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

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

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

Plug in both potential values.

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

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

$x = 9$ $x=9$

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