# Prove $2^{135}+3^{133}<4^{108}$

How can we prove the following inequality?

$$2^{135}+3^{133}<4^{108}$$

• Why don't you just use a calculator and compare the results? – Stefan Hamcke Aug 16 '13 at 16:27
• can he calculate $2^{135}$? – dato datuashvili Aug 16 '13 at 16:32
• @Vijay: If the question was " Prove the inequality $2^4+3^4<4^4$ ", would you still say that using a calculator is an invalid response? I don't think I would. – Eric Stucky Aug 16 '13 at 16:39
• by prove, you mean prove by analytic methods. using a calculator is known as 'verification'-checking your hypothesis – dajoker Aug 16 '13 at 16:41
• @VijayRaghavan: I disagree. Proving is verifying correctness. If explicit calculation can confirm this, then this a perfectly valid proof. On the other hand, this is most likely not what the OP actually wants. – Tomas Aug 16 '13 at 18:40

First,

$$2^{135}=2^7\cdot2^{128}=2^7\cdot(2^8)^{16}<(2^8)^{17}=256^{17}\;.$$

Next,

$$3^{133}=3^3\cdot3^{130}=3^3\cdot(3^5)^{26}=3^3(243)^{26}<3^3(256)^{26}=27\cdot256^{26}\;.$$

Thus,

$$2^{135}+3^{133}<256^{17}+27\cdot256^{26}=(1+26\cdot256^9)256^{17}<256^{10}\cdot256^{17}=256^{27}=4^{108}\;.$$

• Aha! Very nice. – Eric Stucky Aug 16 '13 at 16:43
• Excellent. Way to go!!! – dajoker Aug 16 '13 at 16:44
• Minor simplification: $27\cdot 256^{26}<2^5\cdot2^{8\cdot 26}=2^{213}$, thus $2^{133}+3^{133}<2\cdot 3^{133}<2\cdot 2^{213}=2^{214}=4^{107}$. This is quite close to the "true" exponent $\approx 105.4$. – Hagen von Eitzen Aug 16 '13 at 17:06

While there are several more clever answers up there, I couldn't resist posting this answer.

2^135 =                          43556142965880123323311949751266331066368
3^133 =   2865014852390475710679572105323242035759805416923029389510561523
4^108 = 105312291668557186697918027683670432318895095400549111254310977536


So even by eye, you can confirm that $2^{135}+3^{133}<4^{108}$.

• Since the inequality is so far, and you can do it "by eye", perhaps we could consider using $\log_{10}$ along with estimates for the $\log_{10} 2, \log_{10} 3$ (and then $\log_{10} 4$ follows). – George V. Williams Aug 16 '13 at 19:23
• @GeorgeV.Williams Yeah, sure. – Ali Aug 16 '13 at 19:54
• Hm, one possible concern with eyeballing such long strings is that, without counting, are we sure that the string with less horizontal length has the least number of digits? Depending on the formatting, some characters may take up less space than others. It appears things are pretty equal here though. – Jonas Meyer Aug 16 '13 at 19:59
• @Jonas: how's that? Ali, I hope you don't mind the edit... – Mike F Aug 17 '13 at 0:19
• @JonasMeyer This would be an issue in written form or with usual font but Ali made sure to use a monospace one, which has the same horizontal and vertical space allocation for any kind of symbol. – Glutanimate Aug 17 '13 at 8:27

Using $3^5 < 2^8$ we have $3^{130}<2^{208}$.

$$2^{135}+3^{133}< 2^{135}+3^{3}2^{208}<2^{208}+3^{3}2^{208}=(1+27)4^{104}<4^4 \cdot 4^{104}$$

Note that $2^6<3^4$, so $$2^{135}=2^{129}2^6<2^{129}3^4<3^{129}3^4=3^{133}.$$ Therefore $$2^{135}+3^{133}<3^{133}+3^{133}=2\cdot 3^{133}<3\cdot 3^{133}=3^{134}.$$

Next note that $3^5<2^8$, so $$3^{134}<2^{134\cdot (8/5)}<2^{215}<2^{216}=4^{108}.$$

Hint: log$_2(3)\approx 1.5850$… Consider writing $3^{133}$ as a power of $2$ and factoring.

• It is enough to use $\log_2(3)<8/5$, i.e., $3^5<2^8$. – Jonas Meyer Aug 16 '13 at 16:58

Note that $4=2^2$ and $\frac {3^4}{4^4} \lt 3^{-1}$ - also that $108=4\times 27$

On dividing through by $4^{108}$ and working with the expression we find $$\frac {2^{135}}{4^{108}}+\frac {3^{133}}{4^{108}}=2^{-81}+3^{25}\left( \frac{3^4}{4^4}\right)^{27}\lt 2^{-81} + 3^{-2}\lt \frac 12+\frac 13\lt 1$$

Another one that uses the $$3^5 < 2^8 = 4^4$$

and the easy to prove: $$2^{135} + 3^{133} < 3^{135}$$

Then:

$$2^{135} + 3^{133} < 3^{135} = (3^5)^{27} < (4^4)^{27} = 4^{108}$$

$$3^{135} = 3^{5 \cdot 27} < 4^{4 \cdot 27} = 4^{108} \implies {2^{135}+3^{133}} < 3^{134} < {4^{108}}\;.$$

Take logarithms, and don't worry about even large differences:

$$\begin{eqnarray} 135\log 2&\approx &93.57\\ 133\log 3&\approx &146.12\\ 135\log 2&<&133\log 3\\ 108\log 4&\approx &149.72\\ \log[2^{135} + 3^{133}]&< &\log (2 \cdot 3^{133})\\ \log (2 \cdot 3^{133}) &\approx &146.81 \end{eqnarray}$$

i would like different method,first let us do following things

$2^{135}=2^{27}*2^{108}$

and $3^{133}=3^{25}*3^{108}$

so we have

$2^{27}*2^{108}+3^{25}*3^{108}<4^{108}$

can't proceed from this?

• It is unclear to me what you would suggest doing with this reformulation. – Jonas Meyer Aug 16 '13 at 16:56
• dato, I do not understand what your comment means. If "way" means a way to solve the problem, I don't know what way you are suggesting. There is a reformulation of the inequality, but I don't understand how it is being suggested that this reformulation could be used to prove that it is true. – Jonas Meyer Aug 16 '13 at 17:05
• That is a different inequality, which in the case of $n=108$, is much weaker. It is unclear to me how it would show a solution to this problem. I have no doubt that one could start some solution with this observation, but it does not appear to be indicated how in this answer. – Jonas Meyer Aug 16 '13 at 17:08
• ok if it is bad,you can downvote or i can delete it,no problem – dato datuashvili Aug 16 '13 at 17:11
• @JonasMeyer The solution I posted deals with the $3$ part of the solution in a manner compatible with this, and the $2$ part is easily adaptable. – Mark Bennet Aug 16 '13 at 19:05

$9=3^2>2^3=8 \implies 3^{90}>2^{135}$

\begin{align}256=4^4>3^5=243 &\implies \\ 4^{108} &>3^{135} \\ &>3^{90}+3^{133}\\ &>2^{135}+3^{133}\end{align}

Clearly we are a long way from having a close inequality there. The same approach would justify $2^{201}+3^{134}<4^{108}$