Question 806279
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Whoever "they" are, stop listening to them.  (Or learn to listen better, because if "they" actually knew what they were talking about they most certainly did NOT tell you to square the 3.)


The sum of the logs is the log of the product, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(x)\ +\ \log_2(x\ -\ 2)\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(x^2\ -\ 2x\right)\ =\ 3]


But


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


Hence


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 2x\ -\ 8\ =\ 0]


Factor and solve.  Exclude any negative root (the domain of log is the positive reals).


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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