Question 326418
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The sum of the logs is the log of the product.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(2x\ +\ 5)\ +\ \log(x\ +\ 1)\ =\ \log\left((2x\ +\ 5)(x\ +\ 1)\right)\ =\ \log(2x^2\ +\ 7x\ +\ 5)]


Use the definition of the logarithm function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_b(x) \ \ \Rightarrow\ \ b^y = x]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1 = \log(2x^2\ +\ 7x\ +\ 5)\ \ \ \Rightarrow\ \ 10^1\ =\ 2x^2\ +\ 7x\ +\ 5]


Solve the quadratic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 7x\ -\ 5\ =\ 0]


Exclude any roots that would make *[tex \LARGE 2x\ +\ 5] or *[tex \LARGE x\ +\ 1] outside the domain of the log function.  That is to say, for either of the roots of *[tex \LARGE 2x^2\ +\ 7x\ -\ 5\ =\ 0] the following criteria must be met: *[tex \LARGE 2x\ +\ 5\ >\ 0] and *[tex \LARGE x\ +\ 1\ >\ 0].


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