Question 382269
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\,\cdot\,\log(3)\ +\ \log(2x)\ -\ \log(x\ +\ 1)]


Use 


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


to write:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(9)\ +\ \log(2x)\ -\ \log(x\ +\ 1)]


Use "The sum of the logs is the log of the product"


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(18x)\ -\ \log(x\ +\ 1)]


Use: "The difference of the logs is the log of the quotient"


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\frac{18x}{x\ +\ 1}\right)]


And that, as they say, is that.


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