Question 235044
<font face="Garamond" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x\ -\ 3)\ +\ \log(x)\ =\ 1]


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


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


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x^2\ -\ 3x)\ =\ 1]


If the base is unspecified, base 10 is understood, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_{10}(x^2\ -\ 3x)\ =\ 1]


Apply the definition of logarithms


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ =\ \log_{10}(x^2\ -\ 3x)\ \Rightarrow\ \ 10^1\ =\ x^2\ -\ 3x]


Put that into standard form and you have a factorable quadratic equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 3x\ -\ 10\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ - 5)(x\ +\ 2)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 5] or *[tex \LARGE x\ =\ -2]


But *[tex \LARGE -2] is not in the domain of *[tex \LARGE \log(x)], therefore you must exclude this root.


Check


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(5-3)\ =\ \log(2)\ \approx\ 0.3010]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(5)\ \approx\ 0.6990]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.3010\ +\ 0.6990\ =\ 1] 


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