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- Feb 14, 2012

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Problem:

Solve in the set of real numbers the equation $5^x+5^{x^2}=4^x+6^{x^2}$.

Attempt:

At first glance, we can tell $x=0, 1$ would be the two answers to the problem but how do we prove these two are the only answers?

I think this problem must have something to do with the Mean Value Theorem, because if we let $f(a)=a^x$, we know the function of $f$ is continuous and differentiable on $[4, 5]$, so by the Mean Value Theorem, we have

$f'(c)=\dfrac{5^x-4^x}{5-4}=5^x-4^x$ where $4<c<5$.

Similarly, by letting $g(a)=a^{x^2}$, we know the function of $g$ is continuous and differentiable on $[5, 6]$, so by the Mean Value Theorem, we have

$g'(d)=\dfrac{6^x-5^x}{6-5}=6^x-5^x$ where $5<d<6$.

Recall that we are given $5^x+5^{x^2}=4^x+6^{x^2}$, that means $f'(c)=g'(d)$.

But if we are to differentiate the functions of $f$ and $g$ w.r.t. $a$, we get

$f(a)=a^x$ $f'(a)=xa^{x-1}$ | $g(a)=a^{x^2}$ $g'(a)=x^2a^{x^2-1}$ |

$\therefore f'(c)=xc^{x-1}$ | $\therefore g'(d)=x^2d^{x^2-1}$ |

By equating the two derivatives we get

$xc^{x-1}=x^2d^{x^2-1}$

I don't know how to continue from there hence any help is appreciated!