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# Logarithm help watch

1. y= ln (5x/2x-3)
= ln (5x) - ln (2x-3)
dy/dx= 1/x - 2/(2x-3)

(2x-3)-2x / x(2x-3)
= -3/x(2x-3)

Answer book gives me 3/x(3-2x). What am I forgetting to do?
Thanks
2. (Original post by Moonkin)
y= ln (5x/2x-3)
= ln (5x) - ln (2x-3)
dy/dx= 1/x - 2/(2x-3)

(2x-3)-2x / x(2x-3)
= -3/x(2x-3)

Answer book gives me 3/x(3-2x). What am I forgetting to do?
Thanks
They are the same answer. If you times your answer top and bottom by -1, you get what they have written, it just looks better with a +ve on the top
3. (Original post by 2710)
They are the same answer. If you times your answer top and bottom by -1, you get what they have written, it just looks better with a +ve on the top
Duh, haha wow. Thanks

How would I go about solving

ln (x+3) + 1/(x+3)^2

I did
dy/dx= 1/x+3 +(x-3)^-2

Now eveything I'm doing after this seems wrong
4. (Original post by Moonkin)
Duh, haha wow. Thanks

How would I go about solving

ln (x+3) + 1/(x+3)^2

I did
dy/dx= 1/x+3 +(x-3)^-2

Now eveything I'm doing after this seems wrong
What is the question?
5. (Original post by steve2005)
What is the question?
Differentiate with respect to x

ln(x+3) + 1/(x+3)
6. (Original post by Moonkin)
Duh, haha wow. Thanks

How would I go about solving

ln (x+3) + 1/(x+3)^2

I did
dy/dx= 1/x+3 +(x-3)^-2

Now eveything I'm doing after this seems wrong
You've differentiated a bit wrong. 1/(x+3)^2 does not differentiate into (x-3)^-2
7. (Original post by Moonkin)
Differentiate with respect to x

ln(x+3) + 1/(x+3)

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