Question 380172
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(2x\ +\ 8\right)\ -\ \log_2\left(2x^2\ +\ 21x\ +\ 61\right)\ =\ -3]


First 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)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(\frac{2x\ +\ 8}{2x^2\ +\ 21x\ +\ 61}\right)\ =\ -3]


Then use the definition of the logarithm function:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2x\ +\ 8}{2x^2\ +\ 21x\ +\ 61}\ =\ 2^{-3}\ =\ \frac{1}{2^3}\ =\ \frac{1}{8}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8(2x\ +\ 8)\ =\ 2x^2\ +\ 21x\ +\ 61]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 5x\ -\ 3\ =\ 0]


Then just solve the factorable quadratic.



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