For how many integers k is k^2 = 2^k ?

I have discussed a more generic question on this link: try-this-one-700-level-number-properties-103461.html#p805817

Go through it. It should give you multiple ideas.

Easy, only two ways this can happen 2^2 = 2^2 and 2^4 = 4^2. C

We have to pick numbers,check all multiples of 2 but after 4 none satisfies hence the OA is C.

Take logs on both sides => 2*logk = k*log2. This helps better visualization of the 2 answers 2, 2^2.

Analytically it is not possible to see that these are the only 2 integer solutions. But you can find out that these are the ONLY possible solutions.
For this, TRY GRAPHING

The best way to see that these are the only solutions is to look at the graph of these two functions:
f(k) = kˆ2 is a parabola with vertex on (0,0)
g(k) = 2ˆk is a exponential curve that has (0,1) as its y intersect

If you trace both graphs you'll see that the tail of the exponential curve does in fact cross the parabola for a value of k < 0 . However you can easily cross out the possibility of this being an integer value of k by testing out k = -1 and k = -2. (ps: Just for curiosity ,this value is a non rational value, aprox. -0.76)

For positive values, there are two intersections (that happen to be integers!). They are, in fact, the trivial k=2 and k=4 solutions most people guessed by testing numbers. It is not hard to see that the first intersection will have to happen, and given how the rate of increase for the exponential curve picks up much faster than the rate of increase for the quadratic function, they will have to cross paths once more. After that, the rate of increase for the quadratic does not catch up to the exponential, making it impossible for a third positive intersection.

Hope it helped!

\(4^2 = 2^4\)
\(2^2 = 2^2\)

Answer = C

2^2= 2^2
2^4= 4^2

For k= 2 and 4 the above equality holds true.

C is the answer

Why can we not consider zero in this?

k^2 = 0^2 =0, while 2^k =2^0 = 1.

do it graphically, you can get answer within 20 seconds.
2^k cuts k^2 once when -1<k<0
and twice when k=2 and k=4
Because we need only integral values of k, hence answer is 2.

Hi Bunuel, how can we make sure that our list of 2 integers i.e. 2 and 4 is exhaustive apart from plugging in the numbers. Are their any algebraic ways to solve this?

For how many integers k is k^2 = 2^k ?

(A) None
(B) One
(C) Two
(D) Three
(E) More than three

\(2^k= k^2\) is true for 2 integers:
\(k=2\) --> \(2^2=2^2=4\);
\(k=4\) --> \(4^2=2^4=16\).

Well, \(2^2=2^2=4\) is obvious choices, then after trial and error you'll get \(4^2=2^4=16\) as well. But how do we know that there are no more such numbers? You can notice that when \(k\) is more than 4 then \(2^k\) is always more than \(k^2\) so \(k\) cannot be more than 4. [mk[/m] cannot be negative either as in this case \(2^k\) won't be an integer whereas \(k^2\) will be.

Answer: C.

why can't we also take 1 into consideration? Bunuel

k = 1 does not work: k^2 = 1^2 = 1 while 2^k = 2^1 = 2.