1
$\begingroup$

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?

Thanks,

Kaien

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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

$$y+\frac1y=b,~~y>0$$

$$y^2-by+1=0$$

$$y=\frac{b\pm\sqrt{b^2-4}}2$$

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 $b=2$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ I'll leave the mentions to the comments. $\endgroup$ – Simply Beautiful Art Feb 27 '17 at 1:04
0
$\begingroup$

$a^0 + a^0 = 2$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.