CAT 2023 Slot 3 QA Question 5

The problem says: 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... and the correct answer is 5.

First, We analyze what the problem is asking. It's giving me two inequalities involving n, and We find the smallest integer m that makes the second inequality true for all n that satisfy the first one.

Let me break it down.

The first inequality is 5^{n-1} < 3^{n+1}. We find for which natural numbers n this holds true. Then, for each such n, I have to ensure that 3^{n+1} < 2^{n+m}, and find the minimal m that works for all those n.

So, maybe I should first solve the first inequality to find the range of n. Then, for each n in that range, find the minimal m such that the second inequality holds. Then, among all those m's, pick the largest one because m has to satisfy the inequality for each n, so the maximum m required will be the minimal m that works for all.

Wait, no. Actually, since m has to satisfy 3^{n+1} < 2^{n+m} for each n that satisfies the first inequality, We find the minimal m such that for all n where 5^{n-1} < 3^{n+1}, the second inequality holds.

So, perhaps I should first find the values of n for which 5^{n-1} < 3^{n+1}. Let me solve that inequality.

Starting with 5^{n-1} < 3^{n+1}.

I can rewrite this as (5^{n}) / 5 < 3^{n} * 3.

So, (5/3)^n < 15.

Hmm, because if I divide both sides by 3^{n}, I get (5/3)^n < 15.

Starting with 5^{n-1} < 3^{n+1}.

Divide both sides by 3^{n}:

5^{n-1} / 3^{n} < 3^{n+1} / 3^{n}.

Which simplifies to (5^{n} / 5) / 3^{n} < 3.

So, (5/3)^n / 5 < 3.

Multiply both sides by 5:

(5/3)^n < 15.

Yes, that's correct.

So, (5/3)^n < 15.

Now, I can take the natural logarithm on both sides to solve for n.

ln((5/3)^n) < ln(15).

Which simplifies to n * ln(5/3) < ln(15).

Since ln(5/3) is positive, I can divide both sides by it without changing the inequality direction.

So, n < ln(15) / ln(5/3).

Compute that.

First, compute ln(15). I know that ln(15) is ln(3*5) = ln(3) + ln(5). Approximately, ln(3) ≈ 1.0986, ln(5) ≈ 1.6094. So, ln(15) ≈ 1.0986 + 1.6094 ≈ 2.708.

Then, ln(5/3) is ln(5) - ln(3) ≈ 1.6094 - 1.0986 ≈ 0.5108.

So, n < 2.708 / 0.5108 ≈ 5.3.

Since n is a natural number, n can be 1, 2, 3, 4, or 5.

Wait, but let me check for n=5. Let's plug back into the original inequality.

5^{5-1} = 5^4 = 625.

3^{5+1} = 3^6 = 729.

So, 625 < 729, which is true.

Now, n=6: 5^{5} = 3125, 3^7 = 2187. 3125 > 2187, so n=6 doesn't satisfy the first inequality.

So, n can be 1,2,3,4,5.

Now, for each of these n, We find the minimal m such that 3^{n+1} < 2^{n+m}.

Which can be rewritten as 2^{n+m} > 3^{n+1}.

Taking log base 2 on both sides:

n + m > log2(3^{n+1}) = (n+1) * log2(3).

So, m > (n+1)*log2(3) - n.

Simplify that:

m > n*(log2(3) - 1) + log2(3).

Since log2(3) is approximately 1.58496.

So, log2(3) - 1 ≈ 0.58496.

So, m > n*0.58496 + 1.58496.

Since m must be an integer, m must be greater than the right-hand side. So, for each n, compute the value and take the ceiling to find the minimal m.

But since m has to satisfy this for all n from 1 to 5, We find the maximum of these minimal m's across n=1,2,3,4,5.

So, Computing for each n:

For n=1:

m > 1*0.58496 + 1.58496 ≈ 0.58496 + 1.58496 ≈ 2.16992. So, m must be at least 3.

Wait, no. Wait, 2.16992, so the next integer is 3.

Wait, but

Wait, 0.58496*1 + 1.58496 = 2.16992. So, m must be greater than 2.16992, so minimal integer m is 3.

Wait, but

3^{1+1} = 9.

2^{1