# College calculus question... Find value of k?

For what value of k does the equation e^2x=k(square-root x) have exactly on solution.

The last half of the equation is k times the root of x, if that was not clear

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• 7 years ago

e^(2x) = k√(x)

We first have to assume that, (e^(2x)) = k√(x), has a unique solution. If that's the case then, (e^2x) / √(x) = k, must have a unique solution as well.

As a result the unique solution for, (e^2x) / √(x), occurs at the minimum value for the curve:

y = e^(2x) / √x

Taking the derivative of the function:

y = e^(2x) / √x

y' = (e^2x) * [ (2√x) - (1 / 2√x) ] / x

y' = (e^2x) * [ (2 / √x) - (1 / 2√(x³)) ]

From here, we solve for x. Doing so will give us:

(e^2x) * [ (2 / √x) - (1 / 2√(x³)) ] = 0

[ (2 / √x) - (1 / 2√(x³)) ] = 0

[ (4x - 1) / √(x³) ] = 0

4x - 1 = 0

x = 1/4

We therefore have:

(e^2x) = k√(x)

k = (e^2x) / √x

Substituting 1/4 for all values of x will give us:

k = (e^1/2) / √(1/4)

k = (e^1/2) / (1/2)

e^(1/2) = e

k = e / 1/2

k = 2e

• 7 years ago

e^(2x) = k√(x)

.......e^(2x)