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

1 Answer
Jan 26, 2016

#x=-3,2#

Explanation:

Given equation is #logx+log(x+1)=log6#

I believe you're familiar with the basics of logarithm to know that there's a general identity, which is #logm+logn=logmn#

Now, we apply this general identity into the main equation, so we get #logx(x+1)=log6#. Sonce it's log on both sides, let's remove that to get
#x(x+1)=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.