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Log Laws

Why is: ln(11y-3) - ln3 - lny simplified as : ln((11y-3)/3y)I understand the divison part but not why it isnt 11y-3/3/y, why do the 3 and y multiply??
Consider 100 divided by 5 divided by 4. 100 divided by 5 is 20, then 20 divided by 4 is 5. However this is the same as as 100 divided by (5 x 4) so it must be the same operation. You can always reassure yourself with a quick example like this which your calculator will verify.
You first have to use the law;
a log (x) = log (x)^a
use that with 'a' being '-1' for the negative signs to take the subject of the log to the power of -1, you should then be able to add them together using the log law;
log (a) + log (b) = log (ab)

Answer:

ln (11y-3) - ln (3) - ln (y) = ln (11y-3) + ln (3)^-1 + ln (y)^-1 = ln (11y-3) + ln (1/3) + (1/y)

=ln (11y-3)/(3y)
Original post by BarnabasB
Why is: ln(11y-3) - ln3 - lny simplified as : ln((11y-3)/3y)I understand the divison part but not why it isnt 11y-3/3/y, why do the 3 and y multiply??


Because

ln⁡(11y−3)−ln⁡3−ln⁡y\ln(11y-3) - \ln 3 - \ln y

is the same as

ln⁡(11y−3)−[ln⁡3+ln⁡y]\ln(11y-3) - [\ln 3 + \ln y]

i.e. ln⁡(11y−3)−ln⁡(3y)\ln(11y-3) - \ln(3y).

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