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

Algebra.Com
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)   (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)   (Show Source): You can put this solution on YOUR website!
if , then
....solve for




shade a part between lines from to the right (exclude point )
RELATED QUESTIONS

log2^3x-5=7 (answered by Alan3354)
log2(3x-2)-log2(x-5)=4 (answered by robertb)
#1. Log2(3x-7)+log2(x+2)=log2(x+1) #2.... (answered by CPhill,ikleyn)
I am cannnot solve this problem no matter how hard I try. Any sort of help would be... (answered by longjonsilver)
log2 3x + log2 5 =log2... (answered by nerdybill)
Using the power law solve the folowing to the nearest hundredth. {{{ 2^(x+2) = 5^(x-1) (answered by Fombitz)
log2^x = log2^5... (answered by themathtutor2009)
what is the solution set to the equation log2(3x+1) +... (answered by nerdybill)
7. log2(x - 2) + 5 = 8 - log2... (answered by stanbon)