TITLE: Prove inequality with powers
QUESTION [5 upvotes]: $4^{79}<2^{100}+3^{100}<4^{80}$
The only thing I can think of is the following:
$\ln(3) = 1.098612288$ and $\ln(4) = 1.38629436112$ so
$\frac {\ln(3)}{\ln(4)} = 0.79248125036$ so
$\frac {79}{100} < \frac {\ln(3)}{\ln(4)} < \frac {80}{100} $
I don't know how to continue (and algebra was never my strong point!!).
Thank you!

REPLY [7 votes]: Set $\alpha := 200 \frac{\log 3}{\log 4} = 158.496...$. Writing all terms as powers of $2$ your inequalities become
$$
2^{158} < 2^{100} + 2^\alpha < 2^{160}.
$$
The first inequality is clearly satisfied (since $\alpha > 158$).
For the second inequality, observe that
$$
2^{160} - 2^{\alpha} = 2^{\alpha}(2^{160 - \alpha} - 1) > 2^{\alpha} > 2^{100},
$$
where the first inequality follows from the fact that $160-\alpha > 1$, hence
$(2^{160 - \alpha} - 1) > 2 - 1 = 1$.