Proof by induction that $2^{n+4}<3^n$ for large n

Question

We have an inequality:

$$2^{n+4}<3^n$$

which holds for large n. How to prove by induction that it holds for large n. I got an idea that I would make infimum of the natural numbers bigger and then prove by induction from that set. So the first term would be let's say 100 instead of 1. But I don't know if this is mathematically correct to move a bound of natural numbers.

Side question: What would be the best way to prove this for large n then.

Answer 1

Well.... if $2^{n+4} < 3^n$ then $2*2^{n+4} < 2*3^n < 3*3^n$...........

So $2^{(n+1) + 4} < 3^{n+1}$ so that's your induction step.

But you need to find a base case.

..... But lets take a more direct approach.

$2^{n+4} < 3^n \iff \log_2 2^{n+4} < \log_2 3^n \iff n+4 < n\log_2 3\iff$

$n-n\log_2 3 < -4\iff n(1-\log_2 3) < -4\iff n > \frac {-4}{1-\log_2 3}=\frac 4{\log_2 3-1}$

We can use a calculator to get that exactly but $2^1 < 3 < 2^2$ and $2^{1.5}=2\sqrt{2}\approx 2.828$ so I'd guess that $\log_2 3 \approx 1\frac 23$ and $\frac 4{\log_2 3-1}\approx 6$.

Let's see. $2^{6+4} = 1024$ and $3^{6} =27^2 = (30-3)(30-3) < 30^2 = 900$. And $2^{7+4} = 2048$ while $3^{7} = 3(30-3)(30-3) = 3*(900 - 180 +9)=3*729> 2100$. So that was a good guess.

If $n \ge 7$ then $2^{n+4} < 3^n$ but if $n \le 6$ then $2^{n+4}> 3^n$.

With a calculator. $\log_2 3 \approx 1.5849625007211561814537389439478$ and $\frac {4}{\log_2 3 - 1} \approx 6.8380451654058191079047610486961$

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Alternatively, in a comment I posted:

Thing to notice: $2^{n+4} = 16*2^n$ and as $3$ is a large base than $2$ it will eventuall outpace $16*2^n$. Indeed $2^{n+4} < 3^n \iff 16*2^{n} < 3^n\iff 16 < (1.5)^n$. The graph of $1.5^n$ as $1.5 > 1$ will be an increasing function and or result holds if $n > \log_{1.5} 16=\frac {\log 16}{\log 1.5}\approx 6.8$. So your result is true for $n \ge 7$. No need for induction IF you know the basics of exponential functions. (i.e. That $b^n; b>1$ is increasing....)

Answer 2

Base case :$n=7$

I chose $7$ as it's the smallest number for which the statement is true, I simply got it using trial and error.

we have $3^{7}>2^{11}$ which is correct as $2^{11}=2048 $ and $3^7=2187$

Assuming the statement is true for $n$, we will try to prove it for $n+1$

For $n+1$ we have:

$3^{n+1}>2^{n+5}$

this can be written as:

$3^n*3>2^{n+4}*2$

we have that $3^n>2^{n+4}$ from assumption and obviously $3>2$ which concludes the induction.

Here is a helpful comment that goes deeply into choosing your base case: https://math.stackexchange.com/a/1361551/653542

Answer 3

Yes, it is mathematically correct.

The first thing you should do is to find the first natural n for which the inequality holds. That can be done easily by treating the statment as an equality and solving for n. Clearly, the statement is not valid for small values n. Once you have that value, use mathematical induction as usual.

Answer 4

Answer to the side question:

The simplest proof, for me, would use Bernoulli's inequality (which is proved by induction).

Indeed $2^{n+4}<3^n\iff \Bigl(\dfrac32\Bigr)^n>16$. Now $$ \Bigl(\frac32\Bigr)^n= \biggl(1+\frac12\biggr)^n>1+\frac n2$$ by Bernoulli's inequality, so it is enough to prove that, for $n$ large enough, $$\frac n2>15\iff n>30.$$