If the maximum value of the function f (x) = log2x in the interval [a, 2A] is twice the minimum value, then a is equal to? ∵2＞1, F (x) = log2x is an increasing function ∴2log2a=log22a． ∴loga2=1． ∴a=2． The answer is like this. What I can't understand is that 2log2a = log22a. Why is there ∥ loga2 = 1? Is there any other way?

If the maximum value of the function f (x) = log2x in the interval [a, 2A] is twice the minimum value, then a is equal to? ∵2＞1, F (x) = log2x is an increasing function ∴2log2a=log22a． ∴loga2=1． ∴a=2． The answer is like this. What I can't understand is that 2log2a = log22a. Why is there ∥ loga2 = 1? Is there any other way?

2log 2a=2lga/lg2
log 2 2a=lg2a/lg2
It shows that 2lga = lg2a, that is, LGA + LGA = LG2 + LGA LG2 = LGA, so LG2 / LGA = 1, log A2 = 1
/Denotes a division sign

Elden Ring Premiere

RELATED INFORMATIONS