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#.

#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?