If $2y = a^x + a^{-x}$, can we find $a^x$?

If $2y = a^x + a^{-x}$, can we find $a^x$?

Question

Given that

$2y = a^x + a^{-x}$

, where

$a > 1$

,

$x > 0$

, prove that

$$\begin{equation*} a^x = y + \sqrt{y^2-1}. \end{equation*}$$

If, further,

$2z = a^{3x} + a^{-3x}$

, prove that

$$\begin{equation*} z = 4y^3 - 3y. \end{equation*}$$

Suggestion

What are we asked to find? Do we need to find $x$ itself?

We could just cube the expression for $a^x$ and then work out $a^{-x}$ and so on. But that seems quite painful. I wonder whether there is a simpler way?

Solution

The question asks about $a^x$, not about $x$ itself. So if we write $u=a^x$, the problem will look a little simpler. As $a>1$ and $x>0$, it follows that $u>1$. Then the question becomes:

Given that $2y=u+1/u$, where $u>1$, prove that $u=y+\sqrt{y^2-1}$. If we multiply both sides of the equation by $u$, we get $2yu=u^2 + 1,$ which rearranges to $u^2-2yu+1=0$. Then using the quadratic formula we get $u=\frac{2y\pm\sqrt{(-2y)^2-4}}{2}=y\pm\sqrt{y^2-1}.$

There are two roots to the equation $2y=u+1/u$, and they will be reciprocals of each other, as replacing $u$ by $1/u$ gives exactly the same equation. So one of the roots will be greater than $1$, the other will be less than $1$. Since we require $u>1$, we need to take the positive sign.

So $a^x=u=y+\sqrt{y^2-1}$.

Another way of arguing which sign to use goes like this.

Since $u>1$, $1/u<1<u$, so $2y=u+1/u<u+u=2u$, and thus $y<u$ or $u>y$. So we have to use the positive sign.

Continuing to write $u=a^x$, we are told that $2z=u^3+\dfrac{1}{u^3}$. If we cube $2y=u+\dfrac{1}{u}$, we obtain $(2y)^3 = \left(u + \frac{1}{u} \right)^3 = u^3+3u+\frac{3}{u}+\frac{1}{u^3} = 2z + 3(2y),$ so $8y^3=2z+6y$. Dividing by $2$ and rearranging gives $4y^3 - 3y = z.$

This is related to the hyperbolic functions. If $a = e$, then $y = \cosh x$ and $z = \cosh 3x$, while in general, since $a^x=e^{x\ln a}$, we have $y=\cosh (x\ln a)$ and $z=\cosh (3x\ln a)$. The second result then shows that $\cosh 3x=4\cosh^3 x-3\cosh x$, which is the same as the corresponding formula for (ordinary) cosines.