Find the definition of function y = ln [a ^ x-k * 2 ^ x] [a > 0 and a is not equal to 1, K ∈ R] Domain of

Find the definition of function y = ln [a ^ x-k * 2 ^ x] [a > 0 and a is not equal to 1, K ∈ R] Domain of

Here we will discuss the value of parameter k
From the meaning of the title: A ^ x-k * 2 ^ x > 0
a^x>k*2^x
Because K belongs to R, we need to discuss it
When K0, take the natural logarithm on both sides at the same time, xlna > LNK + xln2, sort it out to get x * (lna-ln2) > LNK
When a > 2, the domain of X is {LNK / (lna-ln2)), + ∞}; when a = 2, the domain of no solution is an empty set; when a = 2, the domain of no solution is an empty set

Given the function f (x) = ln (AX) / (x + 1) - ln (AX) + ln (x + 1), (a is not equal to 0 and R)
1. Find the definition field of function f (x)
2. Find the monotone interval of function f (x)
3. When a > 0, if there is x such that f (x) ≥ ln (2a) holds, the value range of a is obtained
The first question is settled. The reciprocal is also worked out
But I would like to ask what is the relationship between the classification discussion of monotone interval and the domain of definition
That's how a is classified. According to what?

The derivative is - ln (AX) / (x + 1) ^ 2
The second question is (1) a > 0 domain x > 0

Given function f (x) = 2F '(1) x-ln (x + 1)
Find the analytic expression and monotone interval of function f (x)

Please refer to: from the function f (x) = 2F '(1) x-ln (x + 1), because f' (1) is a definite derivative value and a constant, we can make a, i.e. a = f '(1), so f (x) = 2aX ln (x + 1), f' (x) = 2a-1 / (x + 1), so f '(1) = 2a-1 / 2 = a, deduce a = 1 / 2, so we have: F (x) = x-ln (x + 1), (this is the analytic formula) to get the derivation