Given f (x) = loga (a ^ x-1) (a > 0 and a ≠ 1), solve the equation f (2x) = loga (a ^ x + 1)

f(2x)=loga(a^(2x)-1)
f(2x)=loga (a^x+1)
loga(a^(2x)-1)=loga (a^x+1)
a^(2x)-1=a^x+1
a^(2x)-a^x-2=0
(a^x-2)(a^x+1)=0
A ^ x = 2 or a ^ x = - 1 (rounding off because a ^ x > 0)
That is x = loga2

Let loga (c) and logb (c) be the two roots of the equation x ^ 2-3x + 1 = 0

It can be seen from the meaning of the title
loga(c)+logb(c)=3
Loga (c) logb (c) = 1, i.e
1 / logC (a) + 1 / logC (b) = 3, that is logC (AB) / [logC (a) logC (b)] = 3
1/[logc(a)logc（b）]=1
So logC (AB) = 3
logab（c）=1/3

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