Largest solution of $2\ln(e^x-1)=\ln2+\ln(e^x+3)$ I've been trying to solve this equation: $$2\ln(e^x-1)=\ln2+\ln(e^x+3).$$ I am asked for the equation's largest solution, and I come to a point where I get $\ln(e^x(e^x-4)) = \ln5$. I can't find my way forward. When I took a quick look at the answer, it's given to be $\ln5$, which quite disappoints me, because even though I keep solving, my workout doesn't seem to lead me to such an answer. Does anybody know what I'm doing wrong?
Thanks in advance for any response! 
//Lloyd
 A: Your equation is written as : $ \ln(e^x - 1)^2 = \ln(2e^x + 6)$
The logarithm function is $"1-1"$, so by definition, you get : 
$(e^x - 1)^2 = 2e^x + 6 \Leftrightarrow e^{2x} - 2e^x + 1 = 2e^x + 6 \Leftrightarrow e^{2x} - 4e^x  - 5= 0. $
Set $y = e^x$ and then $y^2 = e^{2x} $ and solve the : $ y^2 - 4y - 5 = 0$, find the solutions for y and then just substitute for $y=e^x$ and you have your solutions.
A: To find the largest solution of $2\ln(e^x-1)=\ln2+\ln(e^x+3)$, use the logarithmic identities $a \cdot \ln(b)=\ln(b^a)$ and $\ln(a) + \ln(b) = \ln(a \cdot b)$ to obtain
$$
\ln\left( (e^x-1)^2 \right) = \ln\left( 2 \cdot (e^x+3) \right).
$$
If you exponentiate both sides, you get
$$
(e^x-1)^2 = 2 \cdot (e^x+3).
$$
Now we can use a trick: We substitute $y = e^x-1$ to get
$$
y^2 = 2 \cdot (y + 4),
$$
which is simply a quadratic equation. Its two solutions are $y_1 = 4$ and $y_2 = -2$. If you substitute back, you obtain
$$
x_1 = \ln(y_1 + 1) = \ln(5) \qquad\text{and}\qquad x_2 = \ln(y_2 + 1) = \ln(-1).
$$
However, only $x_1$ makes sense in the realm of real numbers; so, the only (and therefore largest) solution is $x = \ln(5)$.
