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

Question

How can we prove the following inequality?

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

Answer 1

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,

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

Answer 2

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

Answer 3

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?

Answer 4

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

Answer 5

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

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

Answer 6

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 $$\frac {2^{135}}{4^{108}}+\frac {3^{133}}{4^{108}}=2^{-81}+3^{25}\left( \frac{3^4}{4^4}\right)^{27}\lt 2^{-81} + 3^{-2}\lt \frac 12+\frac 13\lt 1$$

Answer 7

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

Answer 8

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

Answer 9

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

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

Then:

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

Answer 10

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

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

Answer 11

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

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

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