Question 376752
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The regular rules of algebra don't get suspended just because you are working with logarithms.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(\log(x)\ +\ \log\left(x^2\ +\ 1\right)\right)\ -\ 3\log(x\ -\ 2)]


First use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(\chi) + \log_b(\upsilon)\ =\ \log_b(\chi\upsilon)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(\log\left(x^3\ +\ x\right)\right)\ -\ 3\log(x\ -\ 2)]


Then use


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


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\left(x^3\ +\ x\right)^2\right)\ -\ \log\left((x\ -\ 2)^3\right)]


Then use


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


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\frac{\left(x^3\ +\ x\right)^2}{(x\ -\ 2)^3}\right)]


Which satisfies the requirements of the problem.  You could multiply out the two binomials, but that would be a decision  based on the requirements of further calculations.


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