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

Dec 15, 2015

$x = 9$

Explanation:

Use the product rule of logarithms: $\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$

Thus, the expression can be written as

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

Distribute.

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

Raise both sides as the power of $10$.

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

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

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

$\left(x + 10\right) \left(x - 9\right) = 0$

$x + 10 = 0$
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

$x = 9$