$4^{79}<2^{100}+3^{100}<4^{80}$
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!
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 $(2^{160 - \alpha} - 1) > 2 - 1 = 1$.
