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A question from 2007 Jane C4

041087

hi guys,i have a question to ask which say the following
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the marking stated that ln y = −ln(x−1) + 2ln(2x−3) + c,why so?

Reply 1

Notnek

Which part don't you understand? Both of those fractions are in the form:

\dfrac{kf'(x)}{f(x)}

where f'(x) is the derivative of f(x). The integral of this form is given by

k\ln(f(x)) + c

Try differentiating the solution and it may make more sense.

Reply 2

theandyguthrie

You can only take ln of the bottom to be the integral if the top of the fraction is equal to the derivative of the bottom.

So we need to change the constant on the numerators, to be equal to the derivative of the denominator.

For the first fraction the derivative of the bottom is 1.
So we need to multiply by -1 to make it equal.
And then we can take ln. Giving -ln(x-1)

For the second fraction the derivative of the bottom is 2, and the top is 4.
So we need to divide by 2, which means we much multiply by 2 to keep it equal.
So then integral equal 2ln(2x-3)

and then plus c

Reply 3

AndrewJames

dy/ y = lny as it is just 1 /y dy , -1/(x-1)---> -ln(x-1) as it has no powers in the denominator. and 4/(2x-3)-----> 2ln(2x-3) because you do it like normal logs and would have 4ln(2x-3) but you have to divide by the differential of the denominator (2x-3) which in this case is 2 so it becomes 2ln(2x-3) also be aware that you could also move that two to become a power.