# How do you solve logx+log(x+1)=log6?

Jan 26, 2016

$x = - 3 , 2$

#### Explanation:

Given equation is $\log x + \log \left(x + 1\right) = \log 6$

I believe you're familiar with the basics of logarithm to know that there's a general identity, which is $\log m + \log n = \log m n$

Now, we apply this general identity into the main equation, so we get $\log x \left(x + 1\right) = \log 6$. Sonce it's log on both sides, let's remove that to get
$x \left(x + 1\right) = 6$

Expand the left hand side and bring $6$ to the left hand side, we get
${x}^{2} + x - 6 = 0$

I'm sure you know about quadratic equations to understand how I end up with a value of $x$ now.