Let f (x) = 1 + logx3, G (x) = 2logx2, where x ＞ 0, X ≠ 1, compare the size of F (x) and G (x)

Let f (x) = 1 + logx3, G (x) = 2logx2, where x ＞ 0, X ≠ 1, compare the size of F (x) and G (x)

∵ (x) = 1 + logx3, G (x) = 2logx2, ∵ f (x) - G (x) = 1 + logx3-2logx2 = 1 + logx3-logx4 = 1 + logx34. Classification discussion: ① if 1 + logx34 = 0, i.e. x = 43, then f (x) = g (x). ② if 1 + logx34 ＜ 0, i.e. logx34 ＜ - 1, the solution is 1 ＜ x ＜ 43, then f (x) ＜ g (x). ③ if 1 + logx34 ＞ 0, i.e. logx34 ＞ - 1, the solution is x ＞ 43 or 0 ＜ x ＜ 1, then f (x) ＞ g (x) In conclusion: ① when x = 43, f (x) = g (x). ② when 1 < x < 43, f (x) < g (x). ③ when x > 43 or 0 < x < 1, f (x) > G (x)

How to compare log with 4 as the base and 5 as the true number and log with 3 as the base and 2 as the true number?

Median method
log(4)5>log(4)4=1
log(3)2log(3)2

Log is based on 0.5, 6 is true, and the order of 0.5 ^ 6 ^ 0.5 is

6 ^ 0.5 greater than 0.5 ^ 6