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# C3 Math help!

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1. Can someone help me with the question?

I've also shown my answer, but it's completely wrong.

Thank you!
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2. bump
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4. Hey, I've written the equation in ax+by+c=0 form. Hope thats still ok
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5. (Original post by Heisenberg___)
Hey, I've written the equation in ax+by+c=0 form. Hope thats still ok
Thank you for the reply! How did you get y=9/4 and dy/dx? I can't, for the life of me, get that!! I keep getting all the decimal places. ln2x = ln2(0.5) = 0.3xxx right? or does it become 0?
6. (Original post by EconEast)
that's wrong, surely?
the differential of ln2x is 2/x no 1/x
d/dx ln2x is 2(1/2x) = 1/x
7. (Original post by jackien1)
Thank you for the reply! How did you get y=9/4 and dy/dx? I can't, for the life of me, get that!! I keep getting all the decimal places. ln2x = ln2(0.5) = 0.3xxx right? or does it become 0?
No worries! You just stick the value of x=0.5 in carefully; I did the calculation first and then squared it. It works out as 9/4 because ln2(0.5) is ln1 which is zero, so you do 3(0.5) = 3/2 and then square that which is 9/4. Use the chain rule to find dy/dx

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