How to find real roots? $4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$ How to find real roots? 
$$4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$$
I try LMVT but vary difficult
 A: The following argument is admittedly on the intuitive side, using concavity mixed with "flatter than" comparisons.
If $x<0$ then since $x^2>0$ we have
$$5^x-4^x<0<6^{x^2}-5^{x^2}.$$
If $0<x<1$ then in fact $0<x^2<x<1$, and the functions $u^x$ and $u^{x^2}$ are both concave down, with $u^{x^2}$ being flatter between 5 and 6 than is $u^x$ there. So
$$6^{x^2}-5^{x^2}<6^x-5^x<5^x-4^x.$$
Finally if $x>1$ then in fact $1<x<x^2$ and the functions $u^x$ and $u^{x^2}$ are both concave up, with $u^x$ being flatter between 5 and 6 than is $u^{x^2}$ there. So
$$6^{x^2}-5^{x^2}>6^x-5^x>5^x-4^x.$$
If the above (admittedly vague) arguments are right, we have ruled out all possibilities other than $x=0,1$ for the equation $5^x-4^x=6^{x^2}-5^{x^2}$.
A: Trying to piece together the above comments. First, the equation can be rewritten:
$$
\frac{5^x - 4^x}{5 - 4} = \frac{6^{x^2} - 5^{x^2}}{6 - 5}
$$
By LMVT, there exist constants $4 \le \alpha \le 5$ and $5 \le \beta \le 6$
such that:
$$
\begin{align*}
\frac{5^x - 4^x}{5 - 4} &= x \alpha^{x - 1} \\
\frac{6^{x^2} - 5^{x^2}}{6 - 5} &= x^2 \beta^{x^2 - 1}
\end{align*}
$$
Equating them results in:
$$
\begin{align*}
\alpha^{x - 1} &= x \beta^{x^2 - 1} \\
\frac{1}{x}    &= \frac{\beta^{x^2 - 1}}{\alpha^{x - 1}} \\
- \ln x        &= x^2 \ln \beta - x \ln \alpha - \ln \frac{\beta}{\alpha}
\end{align*}
$$
All constants are positive.
But here I'm stuck.
