Let n be any natural number such that $$5^{n-1} < 3^{n + 1}$$. Then, the least integer value of m that satisfies $$3^{n+1} < 2^{n+m}$$ for each such n, is

It is given that $$5^{n-1} < 3^{n + 1}$$, where n is a natural number. By inspection, we can say that the inequality holds when n = 1, 2, 3 4, and 5.

Now, we need to find the least integer value of m that satisfies $$3^{n+1} < 2^{n+m}$$

For, n =1, the least integer value of m is 2.

For, n = 2, the least integer value of m is 3

For, n = 3, the least integer value of m is 4.

For, n = 4, the least integer value of m is 4.

For, n= 5, the least integer value of m is 5.

Hence, the least integer value of m such that for all the values of n, the equation holds is 5.