Posted by **tom** on Thursday, April 4, 2013 at 11:18am.

find the value of k such that the equation e^(2x)=krootx has exactly one solution.

- math -
**Reiny**, Thursday, April 4, 2013 at 12:26pm
consider the two separate functions

y = e^(2x) and y = k√x

To have 1 real root, they must touch each other

that is, they must have a common tangent at their point of contact

so for the first : dy/dx = 2e^(2x)

for the second: dy/dx = (1/2)(k)x^(-1/2)

they must be equal

4e^(2x) = k/√x ---> k = 4e^(2x) √x

from the original: k = e^(2x)/√x

so 4e^(2x) √x = e^(2x)/√x

4√x = 1/√x

cross-multiply

x = 4

then using one of the expressions for k

k = e(2x)/√x = e^8/2

check my arithmetic

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