How do I simplify this expression?

With logarithms, log_b x + log_b y= log_b x*y, so since everything in the problem's numerator is base 3, you can put it into one logarithm and multiply all of the fractions:

log_3 (3/4*4/5*5/6*6/7*7/8*8/9)= log_3 1/3

and...

log_3 1/3 = -1, because 3^-1=1/3, 

you can do the same thing in the denominator, but with base 10:

log 4/5 + log 5/6 + log 6/7 + log 7/8 + log 8/9 + log 9/10= log_10 (4/5*5/6*6/7*7/8*8/9*9/10)= log_10 2/5

so... 

You end up with:

-1/log_10 2/5

Im not sure about how to simplify it from there- you could split the denominator into two logs (log_10 2/5 = log_10 2- log_10 5), but I don't know if that would help.

I assume you mean this

[log₃ (3/4) + log₃ (4/5) + log₃ (5/6) + log₃  (6/7) + log₃ (7/8) + log₃ (8/9)] / [ log (4/5) + log (5/6) + log( 6/7 )+ log (7/8) + log (8/9) + log (9/10)]

log₃ (3/4) + log₃ (4/5) + log₃ (5/6) + log₃  (6/7) + log₃ (7/8) + log₃ (8/9  can be written as

log₃3 - log₃4 + log₃4 - log₃5 + log ₃5 - log₃6 + log₃6 - log₃7 + log₃7 - log₃8 + log₃8 - log₃9  =

log₃3 + [ - log₃4 + log₃4] + [ - log₃5 + log ₃5] + [ - log₃6 + log₃6] + [ - log₃7 + log₃7] + [ - log₃8 + log₃8] - log₃9  =

log₃3 - log₃9   =   1 - 2  = -1

And 

log (4/5) + log (5/6) + log( 6/7 )+ log (7/8) + log (8/9) + log (9/10)  =

log4 - log5 + log 5 - log6 + log 6 - log 7 + log 7 - log 8 + log 8 - log 9 + log 9 - log 10 = 

log4 + [ - log5 + log 5] + [ - log6 + log 6] + [ - log 7 + log 7] + [ - log 8 + log 8] + [ - log 9 + log 9] - log10 =

log 4 - log 10  =   log 4 - 1 

So

[log₃ 3/4 + log₃ 4/5 + log₃ 5/6 + log₃  6/7 + log₃ 7/8 + log₃ 8/9] / [ log 4/5 + log 5/6 + log 6/7 + log 7/8 + log 8/9 + log 9/10] =

-1 / [ log 4 - 1]   =

1 / [ 1 - log 4]  ≈  -2.5887

\(\dfrac{log_3\frac{3}{4}+log_3\frac{4}{5}+log_3\frac{5}{6}+log_3\frac{6}{7}+log_3\frac{7}{8}+log_3\frac{8}{9}}{log\frac{4}{5}+log\frac{5}{6}+log\frac{6}{7 }+log\frac{7}{8}+log\frac{8}{9}+log\frac{9}{10}}\)

=\(\dfrac{log_33-log_34+log_34-log_35+log_35-log_36+log_36-log_37+log_37-log_38+log_38-log_39}{log4-log5+log5-log6+log6-log7+log7-log8+log8-log9+log9-log10}\)= \(\dfrac{log_33-log_39}{log4-log10}\)

=\(\dfrac{1-2}{\log\dfrac{2}{5}}\)

=\(-\log\dfrac{2}{5}\)

Same as CPhill's answer.

Why are you guys adding and subtracting them instead of multiplying? What property allows you to do this?

Because Log(3/4) is the same as: Log(3) - Log(4)!!!!!! Test it if you wish.

Doesnt come out the same on my calculator