Prove that the solutions to $\exp(\sinh(\frac{1}{\ln(x)}))=x$ are reciprocals of each other

Question

Prove that the solutions to $\exp(\sinh(\frac{1}{\ln(x)}))=x$ are reciprocals of each other.

Here's what I did:

I took the natural logarithm of both sides of the equation to get $\sinh(\frac{1}{\ln(x)})=\ln(x).$ Next, I multiplied both sides by the inverse hyperbolic sine to get $\frac{1}{\ln(x)}=\sinh^{-1}(x)\ln(x).$ Then, I multiplied both sides by $\ln(x)$ to get $1=\sinh^{-1}(x)\ln^2(x).$ Next I leveraged a formula for the inverse hyperbolic sine to get $1=\ln(x+\sqrt{x^2+1})\ln^2(x).$ I am stuck at this point.

Answer 1

You made an error rewriting $\sinh(\frac{1}{\ln(x)})=\ln x$. Instead, it should be $\frac1{\ln x}=\sinh^{-1}(\ln x)$, or

$$1=\sinh^{-1}(\ln x )\ln x$$

from which it is easy to see that the reciprocal $\frac1x$ is also a solution, since

$$\sinh^{-1}(\ln\frac1x)\ln\frac1x =\sinh^{-1}(-\ln x)(-\ln x)=\sinh^{-1}(\ln x)\ln x $$

Answer 2

All of that work rewriting the problem is not necessary. All you need to do is show that when you substitute in $1/x$ for $x$, it boils down to the original problem.

\begin{align}\frac1x&=\exp(\sinh(1/\ln(1/x)))\\&=\exp(\sinh(-1/\ln(x)))\\&=\exp(-\sinh(1/\ln(x)))\\&=1/\exp(\sinh(1/\ln(x)))\end{align}

Now just reciprocate both sides and you're done.

Answer 3

If $\exp\left(\sinh\left(\frac1{\log(x)}\right)\right)=x$, then\begin{align}\exp\left(\sinh\left(\frac1{\log(\frac1x)}\right)\right)&=\exp\left(\sinh\left(\frac1{-\log(x)}\right)\right)\\&=\exp\left(\sinh\left(-\frac1{\log(x)}\right)\right)\\&=\exp\left(-\sinh\left(\frac1{\log(x)}\right)\right)\\&=\frac1{\exp\left(\sinh\left(\frac1{\log(x)}\right)\right)}\\&=\frac1x.\end{align}

Answer 4

Observe that the function in the LHS is such that

$$f\left(\frac1x\right)=\frac1{f\left(x\right)}.$$