I'm currently stuck on a problem involving logarithmic equations and would appreciate any help that you could provide.
The equation I'm working on is:
$$
\ln \left(e^x+1\right)-\ln \left(e^x-1\right)=x
$$
I'm supposed to then find the sum of the solutions of the equation.
I've attempted to solve the equation by applying the logarithmic identity:
$$
\ln \left(\frac{a}{b}\right)=\ln (a)-\ln (b)
$$
which gives:
$$
\ln \left(\frac{e^x+1}{e^x-1}\right)=x
$$
From here, I exponentiated both sides using the property:
$$
\begin{aligned}
& e^{\ln (x)}=x, \\
& \text { to get: } \\
& \frac{e^x+1}{e^x-1}=e^x
\end{aligned}
$$
Then, I cross-multiplied and simplified the expression to get:
$$
e^{2 x}-2e^x-1=0
$$
At this point, I tried to use the quadratic formula to solve for $e^x$, but I got stuck with a messy expression that I couldn't simplify.
So, I'm turning to you for help. Can you suggest another method for solving this equation? And once the solution is found, can you also tell me how to find the sum of all its solutions?
Thank you in advance for your help!
Best Answer
So if $y = e^x$, then $y^2 - 2y - 1 = 0$, and completing the square gives $$0 = y^2 - 2y - 1 = y^2 - 2y + 1 - 2 = (y-1)^2 - 2.$$ Hence $$(y-1)^2 = 2$$ or $$y - 1 = \pm \sqrt{2},$$ therefore $$y = 1 \pm \sqrt{2}.$$ Since we stated that $y = e^x$, this means $e^x$ is either $1 - \sqrt{2}$ or $1 + \sqrt{2}$. But the first case is impossible since $e^x > 0$ for any real number $x$, and $1 - \sqrt{2} < 0$.
The rest follows trivially.