# Thread: Evaluation of two natural numbers

1. ## Evaluation of two natural numbers

$\displaystyle \displaystyle $A = {7^{15}} + {15^7}$$
$\displaystyle \displaystyle $B = {6^{16}} + {16^6}$$

Hence
a) $\displaystyle \displaystyle $A < B$$
b) $\displaystyle \displaystyle $A > B$$
c) $\displaystyle \displaystyle $A = B$$

I remember evaluating something like that at school, but when was asked today unfortunately could not produce a reasonable answer

2. Originally Posted by Pranas
$\displaystyle \displaystyle $A = {7^{15}} + {15^7}$$
$\displaystyle \displaystyle $B = {6^{16}} + {16^6}$$

Hence
a) $\displaystyle \displaystyle $A < B$$
b) $\displaystyle \displaystyle $A > B$$
c) $\displaystyle \displaystyle $A = B$$

I remember evaluating something like that at school, but when was asked today unfortunately could not produce a reasonable answer
When I was at school 55 years ago we would have done this by looking at the logarithms (to base 10) of the numbers:

$\displaystyle \log(7^{15}) = 15\log7\approx 12.7,\quad \log(15^7) = 7\log15\approx 8.2,$
$\displaystyle \log(6^{16})=16\log6\approx12.5,\quad\log(16^6) = 6\log16\approx7.2.$

It's clear from the logarithms that $\displaystyle 7^{15}$ is considerably bigger than $\displaystyle 6^{16}$, and the other two numbers are several orders of magnitude smaller than either of these. So A must be quite a lot bigger than B.