Write the following as an algebraic expression of x; sinh(lnx)

Question

So I have a questions asking for $sinh(log_ex)$

sinh(x) = $\frac{e^x-e^{-x}}{2}$

So $sinh(log_x)$ = $\frac{e^{lnx}-e^{-lnx}}{2}$

$\frac{e^{lnx}-e^{lnx^{-1}}}{2}$

= $\frac{x - x^{-1}}{2}$

=$\frac{x-1}{2x}$

However the answer is $\frac{x^2-1}{2x}$

Can't seem to find where we went wrong

Answer 1

$$ \frac{x-x^{-1}}{2} = \frac{x-x^{-1}}{2} \frac{x}{x} = \frac{x^2-1}{2x}$$

Answer 2

You were incorrect when saying $\dfrac{x-x^{-1}}{2}=\dfrac{x-1}{2}$. $$\frac{x-x^{-1}}{2}=\frac{x}{2}-\frac{x^{-1}}{2}$$ $$=\frac{x}{2}-\frac{1}{2x}$$ $$=\frac{x^2}{2x}-\frac{1}{2x}$$ $$=\frac{x^2-1}{2x}$$ Hope I helped.