How do you solve #log(16+2b)=log(b^2-4b)#?

Question

3695 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-22T22:21:43

#{-2, 8}# is the solution set.

Use the property #log(a) = log(b) -> a = b#

#16 + 2b = b^2 - 4b#

#0 = b^2 - 6b - 16#

#0 = (b - 8)(b +2)#

#b = 8 and -2#

Checking in the original equation, both solutions work, hence the solution set of #{-2, 8}#.

Hopefully this helps!