Question #25112
1 Answer
Nov 29, 2017
We have:
#1 ≥ log(cx^2 + c) - log(2x^2 + 2x + 7/2)#
#1 ≥ log((cx^2 +c)/(2x^2 + 2x + 7/2))#
We can reverse the log by raising both sides to
#10^1 ≥ (cx^2 +c)/(2x^2 + 2x + 7/2)#
#10(2x^2 + 2x + 7/2) ≥ cx^2 + c#
#20x^2 + 20x + 35 ≥ cx^2 + c#
#20x^2 + 20x + 35 ≥ c(x^2 + 1)#
#(20x^2 + 20x + 35)/(x^2 + 1) ≥ c#
Does this help?