Given that \(2y = a^x + a^{-x}\), where \(a > 1\), \(x > 0\), prove that \[\begin{equation*} a^x = y + \sqrt{y^2-1}. \end{equation*}\]

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.

If, further, \(2z = a^{3x} + a^{-3x}\), prove that \[\begin{equation*} z = 4y^3 - 3y. \end{equation*}\]

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.