Question 306939
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(x\ +\ 1)\ +\ \log_2(x\ -\ 5)\ =\ 4]


The sum of the logs is the log of the product:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left((x\ +\ 1)(x\ -\ 5)\right)\ =\ 4]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(x^2\ -\ 4x\ -\ 5\right)\ =\ 4]


Definition of logarithims:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4x\ -\ 5\ =\ 2^4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4x\ -\ 21\ =\ 0]


Solve like any other quadratic. (Hint: it factors).  However, you may have extraneous roots.  Check both roots in the original equation paying careful attention to the domain of the logarithm function.


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