Judgment: 1. The function y = a ^ x (a > 0, and a ≠ 1) has the same domain as y = loga (a ^ x) (a > 0, and a ≠ 1), and the function y = 3 ^ (x-1) has the same domain as y = x ^ 3 / X /3. The functions y = 1 / 2 + (1 / (2 ^ x-1)) and y = (1 + 2 ^ x) ^ 2 / (x * 2 ^ x) are odd functions Y = 3 ^ (x-1) has the same range as y = (x ^ 3) / X

Judgment: 1. The function y = a ^ x (a > 0, and a ≠ 1) has the same domain as y = loga (a ^ x) (a > 0, and a ≠ 1), and the function y = 3 ^ (x-1) has the same domain as y = x ^ 3 / X /3. The functions y = 1 / 2 + (1 / (2 ^ x-1)) and y = (1 + 2 ^ x) ^ 2 / (x * 2 ^ x) are odd functions Y = 3 ^ (x-1) has the same range as y = (x ^ 3) / X

a^x>0
So which domain of logarithm is also R
So right
3 ^ (x-1) is Y > 0
The latter y = x ^ 2, but x ≠ 0
So Y > 0
That's right

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