Question

2^x=x^2 what is the value of x

Answer 1

There are three values of x :   - 0.76666029 ..   2  , and 4.
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Given 2ˣ = x²    To solve for x.

   Take logarithms on both sides.

    x Log 2 = 2 Log x           as   Log xⁿ  = n Log x
    
Well to solve the exponential equations is not easy. We can use the graphical means to solve some of this type of questions.

See diagram for the graphs of 2^x and x^2.  

x^2 and 2^x meet at x = 2 and x = 4. 

For x > 4,  2^x is always more than x^2. So there are only two solutions.

x = 2 and x = 4.


Perhaps we can use numerical approximation methods like Newton Raphson or Range Kutta methods.  These are iteration methods.

Another Iteration method:
f(x) = 2^x - x^2
f'(x) = 2^x * Ln 2 - 2 x

f(x) = f(1) + f '(1) * (x - 1) 
      = 1 + (Ln4 - 2) (x - 1) =  - 0.6137 x + 1.6137
We want f(x) = 0.

=> x1 = 1.6137/0.6137 = 2.6

Now substitute x1 = 2.6 and find 
  f(x) = 0 = f(2.6) + f '(2.6) * (x2 - 2.6)

Find next value of x as x2.  
Again write the equation  0 = f(x2) + f '(x2) * (x3 - x2)

When the difference between successive values of x is small enough, we have the answer.

Showing the graph on the negative side also, or using the approximation methods on the negative x axis, we get  x = -0.7666602 ..Also at which they are equal.

We have 3 values for x.