Question 326665
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First use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ln(x\ -\ 4)\ -\ 9\ln(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left((x\ -\ 4)^2\right)\ -\ \ln(x^9)]


The difference of the logs is the log of the quotient, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) - \log_b(y) = log_b\left(\frac{x}{y}\right)]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\frac{(x\ -\ 4)^2}{x^9}\right)]


It really depends on where you would want to go from here.  In many cases it would be convenient to leave the expression as it is, but then there are occasions when it might be better to expand the binomial, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\frac{x^2\ -\ 8x\ +\ 16}{x^9}\right)]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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