log2 4 * log4 8 * log8 16 * ……………nth term = 49, what is the value of n?

Key idea: express each logarithm as a ratio of exponents and look for cancellation.

• For the k-th factor (starting at k = 1), the term is log base 2^k of 2^{k+1}, which equals (k+1)/k.

• The product is (2/1) * (3/2) * (4/3) * … * ((n+1)/n).

• All intermediate factors cancel, leaving n + 1.

Set the product equal to 49: n + 1 = 49, so n = 48.

Answer: 48