# log/ln question mathsWatch

Thread starter 1 week ago
#1
: 2ln(...) - 2ln(...) = 2ln3

In the mark scheme, to solve for the unknown (in the brackets, too lazy to type it out), you had to use the log rule of minusing logs means dividing by that log, but I kept it as a minus and when I solved I got a completely different answer. Is there error in my working out, as in keeping the logs minused instead of divided should still work? Or is there a rule or something that I’m not aware of?
Thank you!
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1 week ago
#2
Let's say it's 2ln(x)-2ln(y)=2ln(3). Firstly 2ln(3)=ln(9). So we have ln(x^2)-ln(y^2)=ln(9). Then by the subtraction we get: ln(x^2/y^2)=ln(9). Now we can get rid of the logs and get the equation: x^2/y^2 = 9. There are many solutions to this such as x=3, y=1; or x=6, y=2, etc.
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1 week ago
#3
This is a rule:

ln(X/y) = ln(X)-ln(y)
ln(xy)= ln(X) + ln(y)

https://mathinsight.org/logarithm_basics
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