Question 749560
For {{{log(4,(x+3))}}} and {{{log(4,(x+2))}}}to exist, it must be that {{{x+3>x+2>0}}} <--> {{{x>-2}}}
(because logarithm exist only for positive numbers).
{{{log(4,(x+3))-log(4,(x+2))>=3/2}}} --> {{{log(4,((x+3)/(x+2)))>=3/2}}} --> {{{(x+3)/(x+2)>=4^(3/2)}}} --> {{{(x+3)/(x+2)>=(2^2)^(3/2)}}} --> {{{(x+3)/(x+2)>=2^((2*3/2))}}} --> {{{(x+3)/(x+2)>=2^3}}} --> {{{(x+3)/(x+2)>=8}}}
Since {{{x+2>0}}}, multiplying both sides times {{{(x+2)}}} does not require flipping the inequality sign, so
{{{(x+3)/(x+2)>=8}}} --> {{{x+3>=8*(x+2)}}} --> {{{x+3>=8x+16}}} --> {{{3-16>=8x-x}}} --> {{{-13>=7x}}} --> {{{-13/7>=x}}}
The solution is {{{highlight(-2<x<=-13/7)}}}