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C3 June 2007 question Watch

1. 8(ii)
Find the exact value of the gradient of the curve y = 4lnx-3/4lnx+3 at the point where it crosses the x axis (dy/dx = 24/ x(4lnx+3)^2 )

The first mark on the mark scheme is for "Identifying lnx=3/4" and then x = e^3/4 which would then lead me to the answer. How is the above identified?
2. At the x axis y=0
0 = 4lnx - 3 (multiply by denominator)
3/4 = ln x (add 3, divide by 4)
ln x = 3/4

then
e^(ln x) = e^(3/4) (raise e^)
x = e^(3/4) (e^(ln x) = x)

Hope that helps

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Updated: December 13, 2010
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