How to prove that $7^{31} > 8^{29}$ How can I prove that $7^{31}$ is bigger than $8^{29}$?
I tried to write exponents as multiplication, $2\cdot 15 + 1$, and $2\cdot 14+1$, then to write this inequality as $7^{2\cdot 15}\cdot 7 > 8^{2\cdot 14}\cdot 8$. I also tried to write the right hand side as $\frac{8^{31}}{8^2}$.
 A: Others may bristle at this "proof," but:
$$7^{31} = 157,775,382,034,845,806,615,042,743 \\ 8^{29} = 154,742,504,910,672,534,362,390,528$$
If all else fails, just calculating the expressions and comparing them will work.  This particular problem is only mildly tedious to attack this way if you have pen/paper.
A: Here's an idea for a proof that uses some, but hopefully not too much, arithmetic. If you know the power series
$$-\ln(1-x)=x+{1\over2}x^2+{1\over3}x^3+\cdots$$
then you can get started by finagling the desired inequality as follows:
$$\begin{align}
7^{31}\gt8^{29}&\iff31\ln7\gt29\ln8\\
&\iff31\ln(8-1)\gt29\ln8\\
&\iff31\left(\ln8+\ln\left(1-{1\over8} \right) \right)\gt29\ln8\\
&\iff2\ln8\gt-31\ln\left(1-{1\over8} \right)\\
&\iff6\ln\left(1-{1\over2} \right)\gt-31\ln\left(1-{1\over8} \right)\\
&\iff6\left({1\over2}+{1\over8}+{1\over48}+{1\over64}+\cdots \right)\gt31\left({1\over8}+{1\over128}+{1\over1536}+\cdots \right)
\end{align}$$
The final ingredient is to use the inequality
$${1\over n}x^n+{1\over n+1}x^{n+1}+\cdots\lt{1\over n}\left(x^n+x^{n+1}+\cdots \right)={x^n\over n(1-x)}$$
in truncating the infinite sum on the right.  It may take a couple of attempts to find truncations that work.
Added later (after seeing math110's answer):  I had quite forgotten my own answer (from two years ago) to the problem of proving $\sqrt7^\sqrt8\gt\sqrt8^\sqrt7$.  In it, I showed all the steps necessary to establish
$$-\ln\left(1-{1\over8} \right)\lt{137\over1024}\quad\text{and}\quad6\ln2\gt{1063\over256}$$
So all that remains here is to note that
$$4\cdot1063=4252\gt4247=31\cdot137$$
Whew!
A: The expression
$$7^{31}>8^{29}$$
Is equivalent to
$$31\ln(7)>28\ln(8)$$
where $\ln$ denotes the natural logarithm. As $7>e$, this is equivalent to
$$\frac{31}{28}>\frac{\ln(8)}{\ln(7)}$$
The above relation can then be easily verified by calculator.
Alternatively, along a similar vein
$$7^{31}=\left(7^{\frac{31}{29}}\right)^{29}$$
As $7^{\frac{31}{29}}\approx8.01>8$ (via my pocket calculator), the inequality follows.
Basically I am still showing this through computation, I'm just trying to make the computations a bit nicer.
A: We have to prove $(\frac {7}{8})^{29}>\frac {1}{49}$. Write
$(\frac {7}{8})^{29}=(1-\frac 18)^{29}=[1- (\frac 18)^{29}]+\binom {29} {1}[1-(\frac 18)^{28}]+….+\binom {29} {14}[1-(\frac18)^{14}]$
Just the first term in this sum of positive is already greater than $\frac {1}{49}$. 
One has $[1- (\frac 18)^{29}]>\frac {1}{49}\iff 8^{29}-1 > \frac {8^{29}}{49}$ which is quite clear.
A: First, $7^5=16807$ and $2^{14}=16384$. And we have $\dfrac{7^{31}}{8^{29}}=\left(\dfrac{7^5}{2^{14}}\right)^6\times\dfrac78$.
Now, for $x\ge0$ and integer $n\ge0$, we have $(1+x)^n\ge1+nx$, hence
$$\left(\frac{7^5}{2^{14}}\right)^6\ge1+6\left(\frac{7^5}{2^{14}}-1\right)=1+\frac{6\times423}{2^{14}}=1+\frac{3\times423}{2^{13}}=\frac{9461}{8192}$$
Finally,
$$\frac{9461}{8192}\times\frac78=\frac{66227}{65536}>1$$
Therefore,
$$\frac{7^{31}}{8^{29}}>1$$
