You are Here: Home

# A question from 2007 Jane C4 Tweet

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
TSR launches Learn Together! - Our new subscription to help improve your learning 16-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. A question from 2007 Jane C4
hi guys,i have a question to ask which say the following

the marking stated that ln y = −ln(x−1) + 2ln(2x−3) + c,why so?
2. Re: A question from 2007 Jane C4
Which part don't you understand? Both of those fractions are in the form:

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

Try differentiating the solution and it may make more sense.
3. Re: A question from 2007 Jane C4
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
4. Re: A question from 2007 Jane C4
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.