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Here's what I've tried so far.

Obvious solutions for equation $(1)$ are $x_1=0 $ and $x_2=1$

$$(1) \Leftrightarrow 6^x-3^x=13^x-10^x \Leftrightarrow \frac{6^x-3^x}{3}=\frac{13^x-10^x}{3} $$

Let $\,f(t)=t^x \hspace{0.1em},\;x\in \mathbb{R}\;$,$\;f$ meets the conditions of the MVT on $[3,6]$ and $[10,13]$ $$\exists\,\xi_1\in(3,6): f'(\xi_1)=\frac{f(6)-f(3)}{3}\Leftrightarrow \frac{6^x-3^x}{3}=x\xi_1^{x-1}\hspace{0.5em}\text{ and }\hspace{0.5em}\exists\,\xi_2\in(10,13):f'(\xi_2)=\frac{f(13)-f(10)}{3}$$ $$\Leftrightarrow \frac{13^x-10^x}{3}=x\xi_2^{x-1}$$

And I'm trying to get a contradiction on there being more roots but I can't get to it to work. Is this a valid way to solve this? Maybe I'm missing something trivial. It's highschool maths so I there aren't that many more options I have (Bolzano-Weirstrass, IVT, Rolle's and Min Max Value Theorem or the first and second derivative test).

Angelo
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ericforman
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    Similar: https://math.stackexchange.com/q/2722239/42969, https://math.stackexchange.com/q/1949421/42969, https://math.stackexchange.com/q/3341966/42969 – Martin R Dec 24 '22 at 19:12
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    You can compare derivatives of left and right handsides at $x=0,1$... – Bob Dobbs Dec 24 '22 at 23:32
  • There are no more solutions, this sort of function is known as a Dirichlet polynomial and in 1883, Laguerre has proved a theorem that bound the number of real solutions in $x$ by "sign changes" in the coefficients. see this answer for statement and refs of this theorem. – achille hui Dec 25 '22 at 06:42

1 Answers1

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I suggest another way for solving this question, which solves a slightly stronger version of your question.

Your equation can be written as $$ \frac{6^x+10^x}{16^x}=\frac{3^x+13^x}{16^x}\iff \left(\frac{6}{16}\right)^x+\left(\frac{10}{16}\right)^x = \left(\frac{3}{16}\right)^x+\left(\frac{13}{16}\right)^x \\\iff (\cos\theta_1)^{2x}+(\sin\theta_1)^{2x}=(\cos\theta_2)^{2x}+(\sin\theta_2)^{2x}, $$ where $\theta_1=\cos^{-1}\sqrt\frac{10}{16}$ and $\theta_2=\cos^{-1}\sqrt\frac{13}{16}$.

Now consider $g(\theta;x)=(\cos\theta)^{2x}+(\sin\theta)^{2x}$ for $0<\theta<\frac{\pi}{4}$. The 1st-order derivative of this function w.r.t. $\theta$ is $$ g_\theta(\theta;x)=2x\sin\theta(\cos\theta)^{2x-1}\left[(\tan\theta)^{2x-2}-1\right]. $$ Note that for $\theta\in(0,\frac{\pi}{4})$, we have $\sin\theta(\cos\theta)^{2x-1}>0$ and $\tan \theta\in (0,1)$. Hence, the sign of $g_\theta(\theta;x)$ for $\theta\in(0,\frac{\pi}{4})$ is $$ g_\theta(\theta;x)=\begin{cases} <0&,\quad x>1\\ =0&,\quad x\in\{0,1\}\\ >0&,\quad x<1\ \ ,\ \ x\ne 0 \end{cases}. $$ The latter implication means that $g(\theta;x)$ is an strictly increasing function of $\theta$ for $\theta\in(0,\frac{\pi}{4})$ and $x>1$ and an strictly decreasing function of $\theta$ for $\theta\in(0,\frac{\pi}{4})$ and $x\in(-\infty,1)-\{0\}$. Hence, the equation $g(\theta_1,x)=g(\theta_2,x)$ has no solution for $(\theta_1,\theta_2)\in(0,\frac{\pi}{4})^2$, $\theta_1\ne\theta_2$ and $x\in\Bbb R-\{0,1\}$. Solutions can exist only when $x\in\{0,1\}$, for which $$ g(\theta;0)=2\quad,\quad g(\theta;1)=1, $$ which proves a stronger version of your question:

The equation $(\cos^2\theta_1)^x+(\sin^2\theta_1)^x=(\cos^2\theta_2)^x+(\sin^2\theta_2)^x$ for $(\theta_1,\theta_2)\in(0,\frac{\pi}{4})^2$ and $\theta_1\ne\theta_2$ has only two solutions for $x\in\{0,1\}$.

Mostafa Ayaz
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  • I've been going through this solution again and it got me wondering; Where did you get the intuition to use $\cos\theta$? I'm assuming you've used it before but what is the meaning behind it and why is it a ''slightly stronger version'' of the question? – ericforman Jan 29 '23 at 10:48