Given $a$ is a fixed real number that is greater than one, how many real numbers $b$ are there such that the equation $𝑎^{x} + a^{-x} = 𝑏$ has a unique real solution x?

To solve this problem, I tried to plug in $a = 3$. So, I want to find how many reals $b$ there are such that $3^{x} + 3^{-x} = b$ has a unique real solution x.

Setting $3^x$ = $t$, I got that $t + \frac{1}{t} = b$. It follows that $t^2 - tb + 1 = 0$. From this, it can be obtained that $t = \frac{b \pm \sqrt{b^2-4}}{2}$. I set the discriminant equal to zero: $b^{2}-4=0$, so $b = \pm 2$. This means that $t = \pm 1$. Since $t = 3^{x}$ (previously defined), $x$ is real only when $t = 1$.

From the work, doesn't it follow that there is only one real number $b$ (that is when $b$ = 2) such that there is one unique real solution $x$? But the answer I see says there are an infinite values of $b$ such that there is one real solution x.

What am I missing in my analysis?



  • $\begingroup$ For any $x$ then $-x$ will be a solution. So $a^0 + a^0 = 2$ is the only one with one real x = -x solution. However for $b > 0$ they have only one unique positive real solution. $\endgroup$ – fleablood Feb 27 '17 at 2:15

A shortcut:

If $x$ is any solution to $a^x + a^{-x}=b$, then $-x$ is also a solution (since $a^x+a^{-x}=a^{-x}+a^x$).

So the only case in which there can be exactly one solution is if $x$ and $-x$ are the same solution -- or in other words when the solution is $x=0$.

But then $b=a^0+a^{-0}=2$.

  • $\begingroup$ You have a little typo in the first equation: $x^{-x}$ instead of $a^{-x}$. $\endgroup$ – Pantelis Sopasakis Feb 27 '17 at 0:59
  • $\begingroup$ @PantelisSopasakis: Fixed, thanks. $\endgroup$ – Henning Makholm Feb 27 '17 at 0:59

If we let $y=a^x$, this becomes




For there to be one solution, $b^2-4=0$, so $b=\pm2$. Clearly, $b>0$, since $y>0$, so the only option is $b=2$.

Alternatively, one could note that for every $x$ that satisfies the equality, $-x$ satisfies the equality, hence one must have $x=0$, and thus $b=2$.

  • $\begingroup$ For there to be one solution The condition is to have one single positive solution. This reduces to having one single real solution, since the two roots have product $+1$ thus the same sign, yet may still be worth mentioning. $\endgroup$ – dxiv Feb 27 '17 at 0:57
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    $\begingroup$ I'll leave the mentions to the comments. $\endgroup$ – Simply Beautiful Art Feb 27 '17 at 1:04

$a^0 + a^0 = 2$

For $b \ge 2$ then $a^{\log_a b} + a^{-\log_a b} > a^{\log_a b} = b$.

As $a^x + a^{-x}$ is continuous there must be at least one solution $a^x + a^{-x} = b; x > 0$.

If we note that $a^x + a^{-x} $ is strictly increasing (look at the derivative; $a^x\ln {a} - a^{-x}\ln{a} > 0$ as $a^x > a^{-x}$) the positive solution is unique. (Clearly if $x$ is a solution $-x$ is as well.)

Remains to show $a^{x}+ a^{-x} \ge 2$ with equality holding only if $x = 0$. It's irritating but straightforward to prove for positive $y$ that $y + 1/y \ge 2$ ($y + 1/y \ge 2 \iff y^2 + 1 \ge 2y \iff (y^2 - 2y + 1) = (y-1)^2 \ge 0$).

[actually as $a^{x} + a^{-x}$ is increasing if $x > 0$ and decreasing on $x < 0$, this result is automatic.)

So $b < 2$ has no solution. $b = 2$ has solution $x = 0$. and for $b > 2$ there will be two solutions, one positive, one negative, of equal magnitude.


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