Prove $2^{135}+3^{133}<4^{108}$ How can we prove the following inequality?
$$2^{135}+3^{133}<4^{108}$$
 A: Hint: log$_2(3)\approx 1.5850$… Consider writing $3^{133}$ as a power of $2$ and factoring.
A: 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 $$\frac {2^{135}}{4^{108}}+\frac {3^{133}}{4^{108}}=2^{-81}+3^{25}\left( \frac{3^4}{4^4}\right)^{27}\lt 2^{-81} + 3^{-2}\lt \frac 12+\frac 13\lt 1$$
A: 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,
$$2^{135}+3^{133}<256^{17}+27\cdot256^{26}=(1+27\cdot256^9)256^{17}<256^{10}\cdot256^{17}=256^{27}=4^{108}\;.$$
A: $$3^{135} = 3^{5 \cdot 27} < 4^{4 \cdot 27} = 4^{108} \implies {2^{135}+3^{133}} < 3^{134} < {4^{108}}\;.$$
A: Another one that uses the $$ 3^5 < 2^8 = 4^4 $$
and the easy to prove: $$2^{135} + 3^{133} < 3^{135}$$
Then:
$$2^{135} + 3^{133} < 3^{135} = (3^5)^{27} < (4^4)^{27} = 4^{108} $$
A: 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 $2^{135}+3^{133}<4^{108}$.
A: Using $3^5 < 2^8$ we have $3^{130}<2^{208}$. 
$$2^{135}+3^{133}< 2^{135}+3^{3}2^{208}<2^{208}+3^{3}2^{208}=(1+27)4^{104}<4^4 \cdot 4^{104} $$
A: 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 $$3^{134}<2^{134\cdot (8/5)}<2^{215}<2^{216}=4^{108}.$$
A: Take logarithms, and don't worry about even large differences:
$$
\begin{eqnarray}
135\log 2&\approx &93.57\\
133\log 3&\approx &146.12\\
135\log 2&<&133\log 3\\
108\log 4&\approx &149.72\\
\log[2^{135} + 3^{133}]&< &\log (2 \cdot 3^{133})\\
\log (2 \cdot 3^{133}) &\approx &146.81
\end{eqnarray}
$$
A: 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?
A: $9=3^2>2^3=8 \implies 3^{90}>2^{135}$
$\begin{align}256=4^4>3^5=243 &\implies \\  
4^{108} &>3^{135} \\
&>3^{90}+3^{133}\\
&>2^{135}+3^{133}\end{align}$
Clearly we are a long way from having a close inequality there. The same approach would justify $2^{201}+3^{134}<4^{108}$
