Ln(9/4) + ln (16/9) - ln (3/1)

1. Dec 31, 2013

939

1. The problem statement, all variables and given/known data

If you have, for example, 2 + 4 - 1, you can get the answer (5), by doing both:

= 2 + (4 - 1)

and,

= (2 + 4) - 1

But the same logic does not work with logs: to get the right answer (4/3) here you must do:

=(ln(9/4) + ln (16/9)) - ln (3/1)

and NOT:

ln(9/4) + (ln (16/9) - ln (3/1))

Why exactly is this so?

2. Relevant equations

ln (9/4) + ln (16/9) - ln (3/1)

2. Attempt at a solution

(ln (9/4) + ln (16/9)) - ln (3/1)

= (ln (4) - ln (3))
= ln (4/3) = correct

(ln (9/4) + (ln (16/9)) - ln (3/1))

= ln (9/4) + ln (16/27)
= ln (144/108) = wrong

Last edited: Dec 31, 2013
2. Dec 31, 2013

scurty

What does that reduce to in lowest terms?

3. Dec 31, 2013

939

Ahhhh, such a stupid mistake. Thanks.

4. Dec 31, 2013

scurty

It happens to everybody, you're welcome.

Rather than simplify at the end you can do it earlier by recognizing that 9 divides into 27 evenly and 4 divides into 16 evenly: $$\frac{9}{4} \cdot \frac{16}{27} = \frac{16}{4} \cdot \frac{9}{27}$$.