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

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
crossmultiply
x = 4
then using one of the expressions for k
k = e(2x)/√x = e^8/2
check my arithmetic
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