1. Exponential Equation.

¤ For what value of k the equation e^(2x)=k(x^0.5) has only one solution?

Help appreciated :-)

Originally Posted by Jeet
In this whole forum can't anyone help me !
Its worth to mention the answer should be independent of x and in terms of e.

2. Is this whole question in itself? Or there are other smaller parts preceding this one? Maybe there are hints given previously...

3. This is the whole question. No hints, nothing :-(

4. Ok, I think I got the answer.

If you sketch a graph of $\displaystyle y= e^{2x}$ and that of $\displaystyle y = k\sqrt{x}$, you'll notice that if they meet at a point only, their gradients are the same.

This means, their derivative are equal.

$\displaystyle e^{2x} = k\sqrt{x}$

$\displaystyle 2e^{2x} = \dfrac{k}{2\sqrt{x}}$

$\displaystyle k = 4\sqrt{x}e^{2x}$

Plugging that in our equation:

$\displaystyle e^{2x} = (4\sqrt{x}e^{2x})\sqrt{x}$

$\displaystyle 1 = 4x$

$\displaystyle x = \dfrac14$

Now, plugging this value of x into the first equation:

$\displaystyle e^{2\frac14} = k\sqrt{\frac14}$

$\displaystyle e^{\frac12} = \frac{k}{2}$

$\displaystyle k = 2\sqrt{e}$