Suppose \(a\neq 0\). Compute \(\log_{2^a} 2^b\) in terms of \(a\) and \(b\).
Compute: log2a2b in terms of a and b.
Using this definition: logrs = t ---> rt = s
log2a2b = x ---> (2a)x = 2b
---> 2ax = 2b
---> ax = b
---> x = b/a
Therefore: log2a2b = b/a