How would I show that $2^{50} < 3^{33}$, without a calculator, and what different methods are there of doing this?
Any help would be much appreciated.
Thanks.
P.S sorry if the tag on this post is wrong. I wasn't sure what to put.
$\begingroup$
$\endgroup$
1
-
4$\begingroup$ Worth remarking that $2^3<3^2\implies 2^{51}=\left(2^3\right)^{17}<\left(3^2\right)^{17}=3^{34}$ So $2\times 2^{50}<3\times 3^{33}$ which is very, very close but not quite good enough. $\endgroup$– luluCommented Sep 11, 2017 at 23:53
Add a comment
|
5 Answers
$\begingroup$
$\endgroup$
1
Note that $$ 3^{34}=(2^3+1)^{17}=2^{51}+17\cdot 2^{48}+C>(2+\frac{17}{4})\cdot 2^{50}>3\cdot 2^{50} $$
-
$\begingroup$ Very neat - and (following lulu) $3^{32} = (2^3+1)^{16} = 2^{48} + 16\cdot 2^{45} + (15\cdot 16/2)\cdot 2^{42} + C > 2^{48} + 2^{49}+ (2^{49}-2^{45}) > 2^{50}$ $\endgroup$– JoffanCommented Sep 12, 2017 at 3:51
$\begingroup$
$\endgroup$
0
Working with small powers of $3$ it isn't hard to see that $$2048=2^{11}<3^7=2187$$
That implies that $$2^{44}<3^{28}$$
But since $$2^6<3^5$$ this is enough.
Note: since, in fact, $2^6<3^4$ this argument proves that $\boxed {2^{50}<3^{32}}$
$\begingroup$
$\endgroup$
1
Method 1: Logarithms
Log both sides (since log is increasing) to get a STP:
$50\log 2<33\log 3$
Then compute with a calculator and solve.
Method 2: Approximation (of logarithms)
To make things simpler, we use log base 2. Then we want to show:
$50<33\log_23$.
Since $2^\frac{1}{2}\approx1.414$, we have $\log_23>>\frac{3}{2}$ (>> means by a decent amount). But then $33\log_23>>49.5$, so it is pretty close. I guess for maximum precision you'd need to use better approximations. (At this point I honestly cannot see how to solve the question without a ton of brute-force calculations)
-
$\begingroup$ Without a calculator. Sorry, I was meant to add that.I'll edit the post. $\endgroup$– L.DavisCommented Sep 11, 2017 at 23:52
$\begingroup$
$\endgroup$
You already received smart solutions to your problem.
Let me try to compare $a=50 \log(2)$ to $b=33 \log(3)$; for that, I shall use the very fast converging series $$\log\left(\frac{1+x}{1-x} \right)=2\left(x+\frac {x^3} 3+O(x^5) \right)$$ For $a$ use $x=\frac 13$ and for $b$ use $x=\frac 12$.
So $$a=\frac{2800}{81}\qquad \text{and}\qquad b=\frac{143}{4}=\frac{2860}{80}>a$$ Using the next term in the expansion, we should get $$a=\frac{8420}{243}\qquad \text{and}\qquad b=\frac{2893}{80}=\frac{8679}{240} > a$$
All of the above has been done by hand.
answered Sep 12, 2017 at 5:44
$\begingroup$
$\endgroup$
0
This would be a comment, but I'm still not allowed to do it.
Actually, one can do better than that:
\[ 3^{34}=3^{2\cdot17}=9^{17}=(1+2^3)^{17}=2^{3\cdot17}(1+2^{-3})^{17}> 2^{51}(1+\frac{17}{8})=2^{51}(3+\frac{1}{8})>2^{51}\cdot3 \] Here I used the well-known inequality \[ (1+x)^p\geq1+px, \] valid for $x>-1$ and $p\geq1$.
Hence, we obtain \[ 3^{33}>2^{51}. \] We could still do better than this though, since \[3^{33}>2^{52}.\] Could anyone prove this last inequality without a calculator?
