Question 454385
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(2\ -\ 2x)\ =\ \ln(x\ +\ 5)\ +\ \ln(x\ +\ 4)]

The sum of the logs is the log of the product:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(2\ -\ 2x)\ =\ \ln(x^2\ +\ 9x\ +\ 20)]

Then

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(\chi_1\right)\ =\ \log_b\left(\chi_2\right)\ \Leftrightarrow\ \chi_1\ =\ \chi_2]

Hence

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ -\ 2x\ =\ x^2\ +\ 9x\ +\ 20]

So

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 11x\ +\ 18\ =\ 0]

Just solve for *[tex \Large x].  Exclude any root that would make the argument of any of the original logarithm functions less than or equal to zero.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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