SOLUTION: I can not seem to figure this question out... log[4](x+3)-log[4](x+2) >= 3/2

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I can not seem to figure this question out... log[4](x+3)-log[4](x+2) >= 3/2      Log On


   

Question 749560: I can not seem to figure this question out...
log[4](x+3)-log[4](x+2) >= 3/2
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
For log%284%2C%28x%2B3%29%29 and log%284%2C%28x%2B2%29%29to exist, it must be that x%2B3%3Ex%2B2%3E0 <--> x%3E-2
(because logarithm exist only for positive numbers).
log%284%2C%28x%2B3%29%29-log%284%2C%28x%2B2%29%29%3E=3%2F2 --> log%284%2C%28%28x%2B3%29%2F%28x%2B2%29%29%29%3E=3%2F2 --> %28x%2B3%29%2F%28x%2B2%29%3E=4%5E%283%2F2%29 --> %28x%2B3%29%2F%28x%2B2%29%3E=%282%5E2%29%5E%283%2F2%29 --> %28x%2B3%29%2F%28x%2B2%29%3E=2%5E%28%282%2A3%2F2%29%29 --> %28x%2B3%29%2F%28x%2B2%29%3E=2%5E3 --> %28x%2B3%29%2F%28x%2B2%29%3E=8
Since x%2B2%3E0, multiplying both sides times %28x%2B2%29 does not require flipping the inequality sign, so
%28x%2B3%29%2F%28x%2B2%29%3E=8 --> x%2B3%3E=8%2A%28x%2B2%29 --> x%2B3%3E=8x%2B16 --> 3-16%3E=8x-x --> -13%3E=7x --> -13%2F7%3E=x
The solution is highlight%28-2%3Cx%3C=-13%2F7%29