The equation of the tangent at P is y=3 x - 2 . Let T be the region enclosed by
the graph of f , the tangent [PR] and the line x= k , between x = −2 and x= k
where − 2< k < 1.
Given that are of T is 2k+4, show that k satisfies the equation k^4-6k^2+8=0.
So I know you have to...
The lines (AC) and (BD) intersect at the point P(3,K)
Show K=1.
AC=(4,2) or (x,y)=(5,2)+s(4,2)
BD=(3,-6) or (x,y)=(1,5)+t(3,-6)
A(1,0), B(1,5), C(5,2), D(4,-1)
[PLAIN]http://img840.imageshack.us/img840/6263/1894869.jpg
Hey guys I need help how to solve this equation... Express your answer to the equation in the form alnb
9e^4x-e^2x=0
This is as far as I got
9e^4x=e^2x
ln(9e^4x)=ln(e^2x)
the answer given in the markscheme is x=1/2ln1/9, x=-1/2ln9, x=ln1/3, a=-1/2 and b=9, x=-ln3 (accept a=-1 and...