Question 1018108: Hello could you please help me simplify: log little3 3/4+ log little3 4/5+ log little3 5/6+ log little3 6/7+ log little3 7/8+ log little3 8/9 / log 4/5+ log 5/6+ log 6/7+ log 7/8+ log 8/9+ log 9/10
Found 4 solutions by Boreal, solver91311, ikleyn, MathTherapy:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! log 3 (3/4)(4/5)...(8/9))---everything cancels except for 3/9, or 1/3, the numerator.
denominator log 3(4/5(5/6)...(9/10) or 4/10 or 2/5.
This is log 3 (1/3)/log 3 (2/5)
log 3 (1/3)=log 3 (1)- log 3 (3)=0-1 or -1.
log 3 2/5 is log 3 2/log 3/5
you are dividing by the last, so you invert and multiply by -1.
-log 3(5)/log 3 (2)
That is log 3 (2)/log 3(5), or log 3 (2/5)
Notice that -log3 (1/3) is -(log3 1-log3 3)= -(-1) or 1. The minus sign inverts the log.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
\ +\ \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:
\ =\ \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:
\ -\ \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  without a specified base means base 10, that is: \ =\ \log_{10}(x)) . Other authorities take to mean the natural or base  logarithm, that is: \ =\ \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 ) means in the context of your question.
Since \ =\ 1) for all positive  , and \ =\ n\log_b(x)) , we can simplify further:
\ -\ \log_3(9)}{\log(4)-\log(10)}\ =\ \frac{1\ -\ 2}{\log(4)\ -\ 1}\ =\ \frac{-1}{\log(4)\ -\ 1})
John
My calculator said it, I believe it, that settles it
Answer by ikleyn(52788)  (Show Source):
Answer by MathTherapy(10552)  (Show Source):
|
|
|
