Find k>1 such that x^2=k^x has exactly two real solutions for x.?

In order for there to be exactly two real solutions, one of the intersection points between y=x^2 and y=k^x will be where the two curves are tangent to each other. That is, both the y values and the y' values will be equal at that point. This means we are solving the simultaneous equations ...

.. x^2 = k^x

.. 2x = (k^x)*ln(k)

Using the former in the latter, we have

.. 2x = x^2*ln(k)

.. (2/x) = ln(k)

.. k = e^(2/x)

Using this in the first equation, we have

.. x^2 = (e^(2/x))^x

.. x^2 = e^2

.. x = e

So, the value of k is e^(2/e), and the tangent point is (e, e^2).

.. k ≈ 2.0870652286345329598

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The other point of intersection is x=-e*ProductLog[1/e] ≈ -0.75694510645758

This is not an answer, just a test, because one answer that was here earlier has now disappeared. I may contact the answerer to see if he/she can enter it again, because I came here just now to award him/her best answer .... but it is gone! A Y!A bug, I guess.

Here is the exact answer that J wrote that has now disappeared. I found the answer quite interesting because I had no idea that the value of k could be given exactly in closed form. When I contacted him he said it had not disappeared for him. Hmmm. Here is J's answer:

Nice problem. After at least an hour working on this I found the exact answer, namely k = e^(2/e) ≈ 2.087065228634533. The solutions for that value of k are e and a number whose approximate value is -0.756945106457583664584. The double root is x=e.

Sorry, but the derivation details are too much for me to type in here.