Let a = log23.9, B = log20.7 and C = 2 find the size of ABC

Let a = log23.9, B = log20.7 and C = 2 find the size of ABC

c=log2 4
Then, the larger the size of 3.9, 0.7 and 4, the larger the whole number
So B

Given that a, B and C are all positive integers and ABC = 8, we prove that log2 (2 + a) + log2 (2 + b) + log2 (2 + C) ≥ 6

It is proved that: ∵ B, C are positive integers, ∵ 2 + a ≥ 22a, 2 + B ≥ 22b, 2 + C ≥ 22c ∵ ABC = 8 ∵ (2 + a) (2 + b) (2 + C) ≥ 22a · 22b · 22c = 88abc = 64 (if and only if a = b = C = 2, the equal sign holds) ∵ log2 (2 + a) + log2 (2 + b) + log2 (2 + C) ≥ log2 (2 + a) (2 + b) (2 + C) ≥ log264 = 6

Given that a > 0, b > 0, a and B are not equal to 1, the range of loga (b) + logb (a) is?

If 2 / 3 of loga > 1, then the value range of a is
I don't understand why a > 1 means 2 / 3 of the loga is < 0?

a> 1, y = loga, X is an increasing function, 2 / 3 of loga

Why is the value range of one-third of loga > 0 (0,1) a detailed process and a good score

loga (1/3) >0
Using the formula of changing bottom
So log (1 / 3) a > 0
log(1/3) a>log(1/3) 1
Because y = log (1 / 3) x is a decreasing function on (0, + ∞)
So 0

Let loga3 / 4 > 1, then the value range of a is

loga(3/4)>1=loga(a)
0

If - 1 < loga3 / 4 < 1, find the range of A

When a > 1, 1 / a < 3 / 4, a > 4 / 3; 3 / 4 < A is a > 4 / 3
When 0 < a < 1, 1 / a > 3 / 4, a < 4 / 3; 3 / 4 > a means 0 < a < 3 / 4
So the value range of a is: a > 4 / 3 or 0 < a < 3 / 4

If loga3 / 7

loga3/7 = (lg3/7)/(lga)
When LGA > 0, i.e. a > 1, Lg3 / 71
When lga0
So the final answer is: 3 / 7 > a > 0 or a > 1

Given the function f (x) = loga [(1 + x) / (1-x)] (a > 0, a ≠ 1), find the value range of X with F (x) > 0

f(x)=log_ a(2/(1-x)-1)
When a ∈ (0,1), f (x) > 0, we obtain 2 / (1-x) - 1 ∈ (0,1) & # 8658; X ∈ (- 1,0);
If a ∈ (1, + ∞), f (x) > 0, then 2 / (1-x) - 1 > 1 & # 8658; X ∈ (0,1)

Given the function f (x) = loga (1 + x) - loga (1-x), find the value range of F (x) > 0

F(x)=loga(1+x)-loga(1-x)
=loga(1+x)/(1-x)>0=loga1
1)00
0