SOLUTION: log(base3)3/4 + log(base3)4/5 + log(base3)5/6 + log(base3)6/7 + log(base3)7/8 + log(base3)8/9 divided by log4/5 + log5/6 + log6/7 + log7/8 + log9/10 Please help me solve. Th

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: log(base3)3/4 + log(base3)4/5 + log(base3)5/6 + log(base3)6/7 + log(base3)7/8 + log(base3)8/9 divided by log4/5 + log5/6 + log6/7 + log7/8 + log9/10 Please help me solve. Th      Log On


   

Question 1087381: log(base3)3/4 + log(base3)4/5 + log(base3)5/6 + log(base3)6/7 + log(base3)7/8 + log(base3)8/9 divided by log4/5 + log5/6 + log6/7 + log7/8 + log9/10
Please help me solve.
This is what I have so far, but I am not sure if I am on the right path.
log(base3) (3/4*4/5*5/6*6/7*7/8*8/9) divided by log (4/5*5/6*7/8*8/9*9/10)
Thank you!
Found 3 solutions by Fombitz, ikleyn, MathTherapy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, you're on the right track.
Do the multiplication,
%283%2F4%29%284%2F5%29%285%2F6%29%286%2F7%29%287%2F8%29%288%2F9%29%289%2F10%29=3%2F10
%284%2F5%29%285%2F6%29%286%2F7%29%287%2F8%29%288%2F9%29%289%2F10%29=4%2F10
So then the whole thing equals,
log%28%280.3%29%29%2Flog%28%280.4%29%29=1.314

Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.
Next step notice that

3%2F4%2A4%2F5%2A5%2F6%2A6%2F7%2A7%2F8%2A8%2F9 = 3%2F9 = 1%2F3    (after canceling the terms in the numerator and denominator)


4%2F5%2A5%2F6%2A6%2F7%2A7%2F8%2A8%2F9%2A9%2F10 = 4%2F10       (after canceling the terms in the numerator and denominator)


Then log%283%2C+%283%2F4%2A4%2F5%2A5%2F6%2A6%2F7%2A7%2F8%2A8%2F9%29%29 = log%283%2C+%281%2F3%29%29 = -1   and


log%28%284%2F5%2A5%2F6%2A7%2F8%2A8%2F9%2A9%2F10%29%29 = log%2810%2C+%284%2F5%2A5%2F6%2A7%2F8%2A8%2F9%2A9%2F10%29%29 = log%2810%2C%284%2F10%29%29 = log%2810%2C%284%29%29-1 = 2%2Alog%2810%2C%282%29%29-1,


so your answer is 

. . . = %28-1%29%2F%282%2Alog%2810%2C%282%29%29-1%29 = 1%2F%281-2%2Alog%2810%2C%282%29%29%29.


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

log(base3)3/4 + log(base3)4/5 + log(base3)5/6 + log(base3)6/7 + log(base3)7/8 + log(base3)8/9 divided by log4/5 + log5/6 + log6/7 + log7/8 + log9/10
Please help me solve.
This is what I have so far, but I am not sure if I am on the right path.
log(base3) (3/4*4/5*5/6*6/7*7/8*8/9) divided by log (4/5*5/6*7/8*8/9*9/10)
Thank you!
You first need to state whether the logs AFTER "divided by" are: