How do you solve #Log3x=log4+log(x+3)#?

1 Answer
Apr 19, 2016

x > 0. The negative solution #x = -12# is inadmissible for the given equation. So there is no real solution.

Explanation:

x > 0.
Use #log a - log b = log (a/b) #

#log ((3x)/(x+3))=log 4#

Som #(3x)/(x+3)=4#

#x = -12#.

As x > 0 for log x, #x = -12# is inadmissible.