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

The correct answer is 5.

Let's analyze the problem step-by-step.

We are given two inequalities involving a natural number $n$ (where $n\in \{1,2,3,...\}$):

1. The first inequality defines the set of permissible values for $n$:
$5^{n-1}<3^{n+1}$

2. The second inequality must be satisfied for all such $n$:
$3^{n+1}<2^{n+m}$

We need to find the least integer value of $m$ that satisfies the second inequality for each $n$ that satisfies the first inequality.

Step 1: Solve the first inequality for $n$

Let's rewrite the first inequality:
$5^{n-1}<3^{n+1}$
Taking the natural logarithm (or log to base 10) on both sides:
$(n-1)\ln \left(5\right)<(n+1)\ln \left(3\right)$
$n\ln \left(5\right)-\ln \left(5\right)<n\ln \left(3\right)+\ln \left(3\right)$
Rearranging the terms to group $n$:
$n\left(\ln \right(5)-\ln (3\left)\right)<\ln \left(3\right)+\ln \left(5\right)$
Since $5>3$, we have $\ln \left(5\right)-\ln \left(3\right)>0$. Dividing both sides by this positive quantity gives:
$n<\frac{\ln \left(3\right)+\ln \left(5\right)}{\ln \left(5\right)-\ln \left(3\right)}$

Using the values $\ln \left(3\right)\approx 1.0986$ and $\ln \left(5\right)\approx 1.6094$:
$n<\frac{1.0986+1.6094}{1.6094-1.0986}$
$n<\frac{2.708}{0.5108}\approx 5.3$

Since $n$ must be a natural number, the possible values for $n$ are:
$n\in \{1,2,3,4,5\}$

Step 2: Solve the second inequality for $m$

We want the inequality $3^{n+1}<2^{n+m}$ to hold for all $n\in \{1,2,3,4,5\}$.

Let's solve for $m$ in terms of $n$:
Taking the logarithm with base 2 on both sides:
${\log }_2\left(3^{n+1}\right)<n+m$
$(n+1){\log }_2\left(3\right)<n+m$
$m>(n+1){\log }_2\left(3\right)-n$

Let $f\left(n\right)=(n+1){\log }_2\left(3\right)-n=n\left({\log }_2\right(3)-1)+{\log }_2\left(3\right)$.
Since ${\log }_2\left(3\right)\approx 1.585>1$, the coefficient of $n$, which is ${\log }_2\left(3\right)-1\approx 0.585$, is positive. Therefore, $f\left(n\right)$ is an increasing function of $n$.

For the inequality $m>f\left(n\right)$ to hold for all $n\in \{1,2,3,4,5\}$, it must hold for the maximum value in this range, which is at $n=5$.

Let's evaluate $f\left(5\right)$:
$f\left(5\right)=(5+1){\log }_2\left(3\right)-5=6{\log }_2\left(3\right)-5$

Using ${\log }_2\left(3\right)\approx 1.58496$:
$f\left(5\right)\approx 6\times 1.58496-5=9.50976-5=4.50976$

Since $m$ must satisfy:
$m>f\left(5\right)\approx 4.51$
and $m$ is an integer, the least integer value of $m$ is $5$.