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

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Answer 1 · 2016-01-27T20:21:31

#x=-12#

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