1. Let a,b be real numbers with a, b > 1. Solve: logb (x^a)=(logb x)^a

2. Let a, b be real numbers with a, b > 1. Solve: loga x + logb x = 1
3. Let a, b be real numbers with a < b. Simplify: log2^a (2^a+1)*log2^a+1 (2^a+2)*log2^a+2 (2^a+3).....log2^b-1 (2^b)

log_b (x^a) = (log_b x)^a
a log_b(x) = (log_b x)^a

Let y = log_b (x)

a y = y^a
a = y^(a-1)
y = a^(1/(a-1))
x = b^(a^(1/(a-1))

There exists a separate solution for each a,b > 1.

2. log x / log a + log x / log b = 1
log x (1 / log a + 1 / log b) = 1
log x = (log a log b) / (log a + log b)
log x = log a log b / log ab
x = e^(log a log b / log ab)

3. log_2^a (2^a+1) * log_(2^a+1) (2^a+2) * log_(2^a+2) (2^a+3).....log2^b-1 (2^b)

= log (2^a + 1)/log(2^a) * log(2^a+2)/log(2^a+1) ... * log(2^b)/log(2^b-1),

Which telescopes to:

log (2^b) / log(2^a) = b log 2 / a log 2 = b/a

The answer is b/a.