take for example

① t =3 power of 2 t=log2 3

② t =5 power of 2 t=log2 5

③ The power of t = 14 of 7 is t=log7 14.

So as long as you understand this concept, it is not difficult to do this problem.

You can directly find the power of (8/ 10) t = 1/5 to get t=log8/ 10 1/5, and simplify it to get t=log4/5 1/5.

(8/ 10) =3/5 to the power of t gives t=log8/ 10 3/5. Simplify it and get t=log4/5 3/5.

∫0 < 4/5 < 1∴t = log4/5x is a monotonic decreasing function on (0, +∞), that is, the larger x is, the smaller t is.

The power from ∫y =(8/ 10) to t is simply reduced on R. The bigger X, the smaller T, and the bigger ∴x, the bigger Y.

That is to say, the larger the function x such as t=log4/5 x, the greater the value of the power of (8/ 10).

∴log4/5 1/5 ≤ t < log4/5 3/5

This problem is a combination of numbers and types. Understand this. Doing this type of finale will definitely help more or less.