Question 1435: What is x^2 = 2^x.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! First of all, you should know how to use standard notation about square or power.
If you are in high school level and have learn calculus then it seems
that you have to solve the given equation by view or basic idea.
Direct way: it is clear that if a, b > 0 then a^b = b^a.
Hence, if x = 2, then 2^x = x^2 = 2^2.
So, x = 2 is a root of 2^x = x^2.
Formal proof using logarithm and calculus as below.
2^x = x^2...(*)
Apply log2 on both sides of(*),we have log2 x^2 = log2 2^x,
so 2 log2 x = x log2 2 = x,
Let y = log2 x , we get 2 y = 2^y .
Apply log2 on both sides,we have log2(2y) = log2 2 + log2y
= 1 + log2y = log2 2^y = y.
Let f(y) = log2 y, since f'(y) = 1/(yln2) and f(1) = 0.
By the mean value theorem for f(y) on the interval [1,y] (or [y,1]),
there exists z' between 0 & y such that
(log2y - 0)/ (y -1)= f'(z') = 1/(z' ln2)
So, log2y = (y-1)/ (z' ln2)
Hence, log2y - y + 1 = (y-1)/ (z' ln2) - (y -1) = (y-1)[-1+ 1/ (z' ln2)]
This shows y =1 is a root of the equation log2y - y + 1 = 0.
And,hence 2^y = 2y has a root y=1.
When y = log2 x = 1, satisfies log2y - y + 1 = 0,and hence it also
satisfies 2^y = 2y.
Recall x = 2^y, so x = 2^1 = 2 if y =1.
We see that x =2 is a root of the given equation 2^x = x^2.
Also, f(2) = log2 2 =1.
By the mean value theorem for f(y) on the interval [y,2] ,
there exists z" between y & 2 such that
(log2 y - 1)/ (y -2)= f'(z") = 1/(z" ln2)
So, log2 y -1 = (y-2)/ (z' ln2)
Hence, log2y - y + 1 = log2 y - 1 -y + 2 = (y-2)/ (z' ln2) -(y-2)
= (y-2)[-1 + 1/ (z" ln2)]
Thus,y = 2 satisfies log2y - y + 1 = 0.
Similarly, x= 2^y = 4 is also a root of the given equation 2^x = x^2.
Answewr: Two roots 2 and 4.
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