Question 201009
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Given the same base, the sum of the logs is the log of the product, and if two logs to the same base are equal, then their arguments must be equal.


Symbolically,


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


and


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


So:


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


and therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x^2 + 2x - 3)= \ln(5)]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + 2x - 3 = 5]


Solve the quadratic by ordinary methods -- this one will factor neatly.


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