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

Why did the exponent go to the party alone? Because it couldn't find a root to bring along! Now, let's solve this equation and find the value of k.

To have exactly one solution, the equation e^(2x) = k√x needs to represent a parabola that just touches the x-axis. That means it must have a vertex right on the x-axis.

The vertex of a parabola in the form f(x) = ax^2 + bx + c is given by (-b/2a, f(-b/2a)).

In our case, f(x) = e^(2x) - k√x. So the vertex is at (-(-b/2a), f(-b/2a)) = (b/2a, f(b/2a)).

In this equation, a = 1, b = 0, and c = -k.

So, the vertex is (0/2, f(0/2)) = (0, f(0)) = (0, e^(2*0) - k√0) = (0, 1 - 0) = (0, 1).

To have exactly one solution, the parabola needs to touch the x-axis at exactly one point, which means the vertex must coincide with that point. In other words, (0,1) must be the desired solution.

Since (0,1) is on the curve, plugging in the values into the original equation gives us e^(2*0) = k√0. Simplifying it, we get 1 = k*0, which implies k can be any value.

So, my friend, the value of k can be any real number you want! Be bold and pick any number that tickles your mathematical fancy!

To find the value of k such that the equation e^(2x) = k√x has exactly one solution, we need to examine the behavior of the function and find the condition for a single solution.

Let's analyze the equation:

e^(2x) = k√x

To simplify the equation, we can rearrange it as follows:

e^(2x) / √x = k

Now, let's analyze the behavior of the left-hand side of the equation, e^(2x) / √x, in order to find the conditions for a single solution.

1. As x approaches negative infinity (-∞), the value of e^(2x) / √x approaches 0.
2. As x approaches positive infinity (∞), the value of e^(2x) / √x approaches positive infinity (∞).

Therefore, there must be at least one solution for k.

To find the value of k for exactly one solution, we need to find the point at which the function e^(2x) / √x transitions from having multiple solutions to having just one solution.

To do this, we can take the derivative of e^(2x) / √x with respect to x:

d/dx (e^(2x) / √x)

Using the quotient rule and simplifying:

(e^(2x) * d/dx (√x) - √x * d/dx (e^(2x))) / (√x)^2

(e^(2x) * (1/2√x) - √x * 2e^(2x)) / x

Simplifying further:

(e^(2x)/2 - 2√x * e^(2x)) / x

Next, we need to set this derivative equal to 0 to find critical points:

(e^(2x)/2 - 2√x * e^(2x)) / x = 0

Multiplying both sides by x to eliminate the fraction:

e^(2x)/2 - 2√x * e^(2x) = 0

Combining like terms:

e^(2x)/2 = 2√x * e^(2x)

Dividing both sides by e^(2x):

1/2 = 2√x

Simplifying:

√x = 1/4

Squaring both sides:

x = (1/4)^2

x = 1/16

Now, let's substitute x = 1/16 back into the original equation to find the corresponding value of k:

e^(2*(1/16)) = k√(1/16)

e^(1/8) = k√(1/16)

Simplifying:

e^(1/8) = (k/4)

Multiplying both sides by 4:

4e^(1/8) = k

Therefore, the value of k that allows the equation e^(2x) = k√x to have exactly one solution is k = 4e^(1/8).

To find the value of k such that the equation e^(2x) = k√x has exactly one solution, we need to consider the conditions for which there is only one intersection point between the curve of e^(2x) and the curve of k√x.

Let's break it down step by step:

1. Rewrite the equation:
e^(2x) = k√x

2. Square both sides of the equation to eliminate the square root:
(e^(2x))^2 = (k√x)^2
e^(4x) = k^2x

3. Apply the laws of exponents:
e^(4x) = e^(ln(k^2) * x)

4. Since the base e is the same on both sides, we can equate the exponents:
4x = ln(k^2) * x

5. Divide both sides by x (assuming x ≠ 0, as division by zero is undefined) to isolate k:
4 = ln(k^2)

6. Take the exponential of both sides to eliminate the natural logarithm:
e^4 = k^2

7. Take the square root of both sides to solve for k:
k = ±√(e^4)

Therefore, the value of k such that the equation e^(2x) = k√x has exactly one solution is k = √(e^4) or k = -√(e^4).

Please note that the value of k can be either positive or negative to satisfy the requirement of the given equation having a single solution.

Thanks

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