How would i get 2 from this algebraic equation???

Question

I have this question, 2^x = x ^2
and i know that the value of x is "2".
But how can i get that "2" mathematically???!!!

N.B
^ sign stand for "RISE POWER"

Answer 1

Lots of good ideas there. just to be different, I will say that if you were to do it as a geometric puzzle it becomes what square of area x^2 is equal in area to a square of area 2^x.
This would allow us to use logic to say that the x from the square with area 2^x is equal to 2^2 and therefore so is x^2 equal to 2^2.
I am purposely trying to be a little out there....but you can never know to much geometry, and anything squared...well...it's square.

Answer 2

Look at the graphs of y = 2^x and y = x^2

Clearly they intersect on either side of the y axis. Trying values for both expressions for positive values of x gives
x 1 2 3 4 5 6
2^x 2 4 8 16 32 64
x^2 1 4 9 16 25 36

Thus we have two solutions, 2 and 4, and it is clear that 2^x is going to exceed x^2 for all values of x greater than 4.

For negative values we have

x -1 -1/2
2^x 0.5 0.707 approx
x^2 1 0.25

Clearly there is a point between -1 and -1/2 where the two expressions have the same value ... but I just looked and saw that falzoon has preceeded me, and I've nothing to add, except that you're not quite right in referring to this as an "algebraic equation". It's actually transcendental, and can't be solved by algebraic methods.

Answer 3

There are actually 3 answers to : 2^x = x^2.
They are 2 , 4 and approximately -0.7666646959621231.

You cannot get them by algebraic means, only by
approximation methods, one of which is graphing.

I got them by drawing a graph of y = 2^x - x^2 and then
testing the visual zeros of 2 and 4.

2^2 = 2^2 and 2^4 = 4^2.

For the other zero, it looked about -0.7, so I plugged that
into Newton's iteration formula to get a more exact value.

Answer 4

Taking log on both sides

x*log2 = 2*logx
x / logx = 2 / log2

comparing both sides
x = 2

Answer 5

take the log of both sides
Log (2^x) = X*log2
Log(X^2) = 2*LogX

X/ (LogX) = 2/ Log 2
*you may assume that the same structure here is self-explainitory