SOLUTION: log2(3x-5)>log2(x+7) I'm so stumped. Any help would by appreciated.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: log2(3x-5)>log2(x+7) I'm so stumped. Any help would by appreciated.       Log On


   

Question 739759: log2(3x-5)>log2(x+7)

I'm so stumped. Any help would by appreciated.
Found 2 solutions by Edwin McCravy, MathLover1:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
log2(3x - 5) > log2(x + 7)

Since logarithms with base greater than 1 are always increasing,
what they are logs of are in the same order of inequality as the
logs themselves are.  Therefore we know that

    3x - 5 > x + 7
        2x > 12
         x > 6 

Edwin

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
if log%282%2C%283x-5%29%29%3Elog%282%2C%28x%2B7%29%29+, then
3x-5%3Ex%2B7+....solve for x
3x-x%3E5%2B7+
2x%3E12
x%3E6

shade a part between lines from x=6 to the right (exclude point x=6)