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

We need to find the range of natural numbers $n$ that satisfy the first condition, then determine the minimum value of $m$ for the second condition.

Let us check which natural numbers satisfy $5^{n-1} < 3^{n+1}$ by testing different values:

For $n = 1$: $5^0 < 3^2$ → $1 < 9$ (true)

For $n = 2$: $5^1 < 3^3$ → $5 < 27$ (true)

For $n = 3$: $5^2 < 3^4$ → $25 < 81$ (true)

For $n = 4$: $5^3 < 3^5$ → $125 < 243$ (true)

For $n = 5$: $5^4 < 3^6$ → $625 < 729$ (true)

For $n = 6$: $5^5 < 3^7$ → $3125 < 2187$ (false)

As $n$ increases, $5^{n-1}$ grows faster than $3^{n+1}$ because the base 5 is larger than 3. Eventually, the exponential with base 5 will overtake the one with base 3.

Therefore, $n$ can be 1, 2, 3, 4, or 5.

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Now we need $3^{n+1} < 2^{n+m}$ to hold for all valid values of $n$ (that's what "for each such $n$" means).

Since we need this inequality to work for ALL valid $n$ values, we should focus on the most restrictive case - the largest value of $n$.

The largest valid $n$ is 5, so let us use $n = 5$:

$3^{5+1} < 2^{5+m}$

$3^6 < 2^{5+m}$

$729 < 2^{5+m}$

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We need to find the smallest integer $m$ such that $2^{5+m} > 729$.

Let us check powers of 2:

$2^9 = 512 < 729$ (not enough)

$2^{10} = 1024 > 729$ (sufficient)

So we need $5 + m \geq 10$, which gives $m \geq 5$.

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Let us check that $m = 5$ works for all other values of $n$:

For $n = 1$: $3^2 < 2^{1+5}$ → $9 < 64$ (true)

For $n = 2$: $3^3 < 2^{2+5}$ → $27 < 128$ (true)

For $n = 3$: $3^4 < 2^{3+5}$ → $81 < 256$ (true)

For $n = 4$: $3^5 < 2^{4+5}$ → $243 < 512$ (true)

All inequalities hold.

Therefore, the least integer value of $m$ is 5.