Question 466404
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The sum of the logs is the log of the product.


Unspecified base implies base 10


Definition of the logarithm function.


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



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


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ +\ 3x\ +\ x^2\ =\ 10^2\ =\ 100]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 3x\ -\ 98\ =\ 0]


Not factorable.  Use the quadratic formula to solve.  Note: The discriminant is prime.  Exclude the negative root because anything less than -1 is excluded from the domain of your most restrictive log argument, i.e., (1 + x).


Use a calculator to check your work using numerical methods.


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