How do you solve #logx-log3 = log(x+8)#?

1 Answer
Jan 27, 2016

#x=-12#

Explanation:

Subtracting logs is the result from taking loges of a division

So #log(x)-log(3) -> log(x/3)#

Write as #color(white)(..)log(x/3)=log(x+8)#

If this is true then it is also true that:

#x/3=x+8#

#x=3x+24-> # Multiplied both sides by 3

#2x=-24->#Collecting like terms and simplifying

#x=-12-># Divide both sides by 2