Solving an Exponent Equation: How to Prove the Two Solutions?

[anemone]

Hi MHB,

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!:)

 

[I like Serena]

Very creative. :)

What you have, is:
$$\text{For every $x$, there is a $4<c<5$ and a $5<d<6$, such that }xc^{x-1}=x^2d^{x^2-1}$$

Now suppose x<0. Then the LHS is negative while the RHS is positive.
So there are no solutions with x<0.

Suppose x>0, then we can divide by x and take the $\ln$ from both sides.
\begin{array}{}
c^{x-1}&=&x d^{x^2-1}\\
(x-1) \ln c &=& \ln x + (x^2-1) \ln d\\
(x-1) ((x+1)\ln d - \ln c) + \ln x &=& 0
\end{array}
Since d > c that means that $(x+1)\ln d - \ln c > 0$.
Now if 0 < x < 1, the LHS is negative, which is a contradiction.
And if x > 1, the LHS is positive, which is again a contradiction.

Therefore $x=0$ and $x=1$ are the only solutions.

 

[anemone]

Very creative. :)

What you have, is:
$$\text{For every $x$, there is a $4<c<5$ and a $5<d<6$, such that }xc^{x-1}=x^2d^{x^2-1}$$

Now suppose x<0. Then the LHS is negative while the RHS is positive.
So there are no solutions with x<0.

Suppose x>0, then we can divide by x and take the $\ln$ from both sides.
\begin{array}{}
c^{x-1}&=&x d^{x^2-1}\\
(x-1) \ln c &=& \ln x + (x^2-1) \ln d\\
(x-1) ((x+1)\ln d - \ln c) + \ln x &=& 0
\end{array}
Since d > c that means that $(x+1)\ln d - \ln c > 0$.
Now if 0 < x < 1, the LHS is negative, which is a contradiction.
And if x > 1, the LHS is positive, which is again a contradiction.

Therefore $x=0$ and $x=1$ are the only solutions.

Awesome, I like Serena! Thank you so much for the easy to understand reasoning to show that those two $x$ values are the only solutions to the problem.

Thank you!