Solving system of logarithmic equations. I have two equations
\begin{align}
\frac{1}{2} \ln(w+b-0.4)+\frac{1}{2}\ln(w-0.4) & =\ln(w)\\
\frac{1}{2} \ln(w+b-0.4)+\frac{1}{2}\ln(w-0.4) & =1.2
\end{align}
I want to find $b$ and $w$.
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I find $w=e^{1.2}$
But when I derive $b$, I obtain that $b= \frac{e^{2.4}}{e^{1.2}-0.4}-e^{1.2}+0.4$
But the value of $b$ Seems not good to me. Please share your answers for the value of $b$. Thank you.

From the first equation, we have
\begin{align}
\frac12\log(w+b-0.4)+\frac12\log(w-0.4)&=\log w\\
\implies\log\left(\sqrt{w+b-0.4}\sqrt{w-0.4}\right)&=\log w\\
\implies\sqrt{(w+b-0.4)(w-0.4)}&=w\\
\implies(w+b-0.4)(w-0.4)&=w^2\\
\implies w&=0.4\left(\frac{5b-2}{5b-4}\right)\\
\implies w&=0.4\left(1+\frac2{5b-4}\right)\tag1
\end{align}
We know that $\log w=1.2\implies w=e^{1.2}$.  Using this to solve for $b$ from $(1)$, we have
$$e^{1.2}=0.4\left(1+\frac2{5b-4}\right)\implies b=0.2\left(\frac{2}{2.5e^{1.2}-1}+4\right)\approx0.854792,$$ and is the same result upon comparison to your answer.
