Quick methods question

[horizon]

Find the value of k, where k is a real number, for which the curve with equation y=loge(x), x >0, intersects the curve with equation y=kx^2 at exactly one point.

Thanks in advance.

[thushan]

Good question. Have no idea how to do this except by trial and error, for which i get k = 0.183932...

[Water]

Hey horizon, first you need to find the derivative of both graphs, for the tangent. To which they will have same derivative.

2kx^2 = log e x

2kx = 1/x

2kx^2 = 1

k = 1/2x^2


After that, you sub this in your main equation


kx^2 = log e x.


And you will have one point, in which it will intersect at one point, based on tangent intersection



By the end , k should equal to 1/ 2e, the same as thusthans. GOod luck

[thushan]

Hey, NICE solution! I did NOT think of that. Respect +1 for you.