yes , f-1(x) = ex - e-x2
for x>0 , ex - e-x =2e-x , ex=√3 ...(1 soln)
for x<0 , ex - e-x = 2ex , e2x = -1/2 (no soln)
so total only 1 soln.
\hspace{-16}$If $\mathbf{f:\mathbb{R}\rightarrow \mathbb{R}}$ and $\mathbf{f(x)=\ln(x+\sqrt{x^2+1})}$\\\\ Then no. of solution of the equation $\mathbf{\mid f^{-1}(x)\mid = e^{-\mid x \mid}}$ is
2?
we can find the inverse by the usual method which comes out to be,
ex-e-x2
so drawing both the graphs we get 2 points
yes , f-1(x) = ex - e-x2
for x>0 , ex - e-x =2e-x , ex=√3 ...(1 soln)
for x<0 , ex - e-x = 2ex , e2x = -1/2 (no soln)
so total only 1 soln.