Question 224352
<font face="Garamond" size="+2">


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


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


So:


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


If two logs to the same base (here, understood base 10) are equal then their arguments must be equal:


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


Hence:


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


Just solve your quadratic.  Check BOTH roots for extraneous values considering the domain of the log function.


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