Compare $4^x+1$ and $2^x+3^x$ for non-negative real $x$ Is it possible to find which on is bigger without calculus?I've thought that since $4^x=2^{2x}$ there could be a quadratic but it doesn't seem right,other then that I tried dividing by $2^x$,$3^x$ but no luck.Also thought about logarithms but they just don't seem to fit here.
 A: Hint: define $f(x)=4^x+1-2^x-3^x$ and show that it has just one positive root at $x=1$. 
A: Let $f(x)=4^x-3^x-2^x+1$.  Note that $f(0)=f(1)=0$.  If you draw the graph, you find that $f$ is strictly negative for $0\lt x\lt1$ and strictly positive (in fact, strictly increasing) for $x\gt1$.  I don't offhand have any good idea for a non-calculus explanation for the function's behavior in $(0,1)$, but here's a way let see why it's strictly increasing (and positive) on $(1,\infty)$.
Let $x=u+1$, and write
$$\begin{align}
f(x)&=4^x\left(1-\left(3\over4\right)^x-\left(1\over2\right)^x+\left(1\over4\right)^x \right)\\
&=4^u\left(4-3\left(3\over4\right)^u-2\left(1\over2\right)^u+\left(1\over2\right)^{2u} \right)\\
&=4^u\left(3\left(1-\left(3\over4\right)^u \right)+\left(1-\left(1\over2\right)^u \right)^2 \right)\\
\end{align}$$
For $u\gt0$, the expressions $1-(3/4)^u$ and $1-(1/2)^u$ are both positive and strictly increasing (to $1$), so the overall expression is also positive and strictly increasing.  It doesn't require calculus to see any of this, just the observation that $3/4$ and $1/2$ are smaller than $1$.
(Note:  $1-(1/2)^u$ is strictly increasing but negative when $u\lt0$, so its square is strictly decreasing for $u\lt0$.  If anyone has a good idea how to handle the range $-1\lt u\lt0$, I'd be happy to see it.)
