SOLUTION: We were asked to solve logarithmic inequations, and although I could solve it, I stopped at a dead end, in which I couldn't find a ''good'' answer.. i) 2*ln(x+2)-1 > 0 ii) log(

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: We were asked to solve logarithmic inequations, and although I could solve it, I stopped at a dead end, in which I couldn't find a ''good'' answer.. i) 2*ln(x+2)-1 > 0 ii) log(      Log On


   

Question 1028184: We were asked to solve logarithmic inequations, and although I could solve it, I stopped at a dead end, in which I couldn't find a ''good'' answer..
i) 2*ln(x+2)-1 > 0
ii) log(3-2x) > 1 + log(x+5)
*Sorry if I posted this again*
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
i) 2ln%28x%2B2%29-1+%3E+0+
==> ln%28x%2B2%29+%3E+1%2F2
==> x%2B2+%3E+e%5E%281%2F2%29+=+sqrt%28e%29
==> x+%3E+sqrt%28e%29+-+2 ≈ -0.35.
ii) log%28%283-2x%29%29+%3E+1+%2B+log%28%28x%2B5%29%29+
==> log%28%283-2x%29%2F%28x%2B5%29%29+%3E+1
==> %283-2x%29%2F%28x%2B5%29+%3E+10
==> %283-2x%29%2F%28x%2B5%29+-+10+%3E0
==> %283-2x-10x-50%29%2F%28x%2B5%29%3E0
==> %28-12x+-+47%29%2F%28x%2B5%29+%3E+0
The critical numbers are -47/12 and -5. (The numbers that make the numerator and denominator equal to zero.)
Choose test numbers -6, -4, and 0. (Corresponding to the partition (-infinity,-5), (-5,-47/12), (-47/12, infinity). )
Upon substitution into %28-12x+-+47%29%2F%28x%2B5%29, the signs are (-), (+), and (-).
Therefore the solution is the open interval (-5,-47/12).