Solve the equation $\operatorname{arcsinh}=\operatorname{arcsech}(x)$ analytically

Question

I am trying to obtain an analytical solution of the equation.

$$\operatorname{arcsinh}(x) = \operatorname{arcsech}(x)$$

Equating the logarithmic definitions leads to the rather unwieldy equation

$$x^4+x^3\sqrt{x^2+1} +x^2 -1.0 -\sqrt{1-x^2}$$

Needless to say I am struggling to obtain an expression for x ! Can anyone offer a solution ?

Answer 1

$$\log(x+\sqrt{x^2+1})=\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)$$

is equivalent to

$$x^2-1=\sqrt{1-x^2}-x\sqrt{x^2+1}.$$

Then with squaring,

$$x^4-2x^2+1=1-x^2-2x\sqrt{1-x^4}+x^2(x^2+1)$$

simplifies to

$$x=0\lor x=\sqrt{1-x^4}.$$

The last equation can be reduced to biquadratic.

$$x=\dfrac1{\sqrt\phi}.$$

Answer 2

$$\sinh ^{-1}(x)=\text{sech}^{-1}(x)$$ $$\sinh\left(\sinh ^{-1}(x) \right)=\sinh\left(\text{sech}^{-1}(x) \right)\implies x=\frac{\sqrt{1-x^2}}{x}$$ Then, after squaring, this reduces to a quadratic in $x^2$.

Answer 3

Let $y={\rm arcsinh}(x)={\rm arcsech}(x)$, then $x=\sinh y={\rm sech}y=1/\cosh y$, so $(e^y-e^{-y})/2=2/(e^y+e^{-y})$, and take it from there.