Question 1018108
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\log_3\left(\frac{3}{4}\right)\ +\ \log_3\left(\frac{4}{5}\right)\ +\ \log_3\left(\frac{5}{6}\right)\ +\ \log_3\left(\frac{6}{7}\right)\ +\ \log_3\left(\frac{7}{8}\right)\ +\ \log_3\left(\frac{8}{9}\right)}{\log\left(\frac{4}{5}\right)\ +\ \log\left(\frac{5}{6}\right)\ +\ \log\left(\frac{6}{7}\right)\ +\ \log\left(\frac{7}{8}\right)\ +\ \log\left(\frac{8}{9}\right)\ +\ \log\left(\frac{9}{10}\right)}]


Start by using the fact that the log of the quotient is the difference of the logs, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left(\frac{3}{4}\right)\ =\ \log_3(3)\ -\ \log_3(4)]


Then if you make all of the appropriate substitutions and collect all like terms (like terms have equal arguments), you end up with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\log_3(3)\ -\ \log_3(9)}{\log(4)\ -\ \log(10)}]


Now there is one little bump in the road at this point.  Some teachers/instructors/professors/mathematicians/textbook authors use the convention that *[tex \Large \log] without a specified base means base 10, that is:  *[tex \Large \log(x)\ =\ \log_{10}(x)].  Other authorities take *[tex \Large \log] to mean the natural or base *[tex \Large e] logarithm, that is: *[tex \Large \log(x)\ =\ \ln(x)\ =\ \log_e(x)].  Since the latter interpretation makes your problem rather messy, I'm going to assume the former convention. However, in the future if you want to communicate with someone who is not familiar with the context in which you are asking the question, then you should either specify the base explicitly or specify at the beginning what *[tex \Large \log(x)] means in the context of your question.


Since *[tex \Large \log_b(b)\ =\ 1] for all positive *[tex \Large b], and *[tex \Large \log_b\left(x^n\right)\ =\ n\log_b(x)], we can simplify further:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\log_3(3)\ -\ \log_3(9)}{\log(4)-\log(10)}\ =\ \frac{1\ -\ 2}{\log(4)\ -\ 1}\ =\ \frac{-1}{\log(4)\ -\ 1}]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>