I want to find $x$ in this mathematical equation

Question

$$0.5=\dfrac{e^{1+x}-e^{-1-x}}{e^{1+x}+e^{-1-x}}$$

How do you find the value of $x$ in the equation if you allow

Answer 1

$$\tanh x={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}$$ so $$\tanh (x+1)={\frac {e^{x+1}-e^{-x-1}}{e^{x+1}+e^{-x-1}}}=0.5$$ $$x+1=\tanh^{-1}0.5$$ $$x=\tanh^{-1}0.5-1$$

Answer 2

Hint. one may just set $X=e^{x+1}$ then solve $$ \frac12=\frac{X-\frac1X}{X+\frac1X} $$ that is $$ X^2+1=2\left(X^2-1\right). $$

Answer 3

your equation is equivalent to $$e^{1+x}+\frac{1}{e^{1+x}}=2e^{1+x}-\frac{2}{e^{1+x}}$$ setting $$u=e^{1+x}$$ we get $$u+1/u=2u+2/u$$ or $$3=u^2$$ can you proceed?