slove : x^2=4^x
x^2=4^x
let's use logs
log (x^2) = log (4^x)
2 log x = x log 4 , not getting anywhere, can't isolate x
graphing it to approximate ....
http://www.wolframalpha.com/input/?i=x%5E2%3D4%5Ex
shows a solution at x = appr -.64
How about Newton's method?
let y = x^2 - 4^x
dy/dx = 2x - (ln4)(4^x)
new x = x - (x^2 - 4^x)/(2x - ln4(4^x))
start with x = -.64
x ... newx
-.64 -.641186
-.641186 -.641185744
-.641185744 -.641185744, wow just 3 steps!!
check:
-.641185744^2 = .411119159
4^-.641185744 = .41119159
x = -.641185744
Just to chime in, in case you haven't studied calculus yet, there's always the bisection method. Let
f(x) = x^2 - 4^x
You want to find where f(x) = 0. Since
f(-1) > 0
f(0) < 0
f(x) = 0 for some x in the interval [-1,0]
so, start guessing at the center of the interval. pick whichever half of the interval the change of sign appears in.
f(-0.75) = 0.2089
f(-0.625) = -0.2982
f(-0.6875) = 0.0871
f(-0.65625) = 0.2804
... 9 steps later ...
f(-0.6412353515625) = 0.000091888
As you can see, it converges rather more slowly than Newton's method. It took 13 iterations to get the same accuracy of 3 steps with Newton's method.
To solve the equation x^2 = 4^x, we need to find the values of x that satisfy the equation. Since the equation involves both a variable (x) and an exponent (4^x), we need to approach this problem using algebraic techniques.
To start, let's rewrite the equation in a more manageable form. Recall that 4 can be written as 2^2, so we have:
x^2 = (2^2)^x
When raising an exponent to another exponent, we multiply the exponents. Applying this property to our equation gives us:
x^2 = 2^2x
Now, we want to simplify further. Taking the square root of both sides of the equation will help us eliminate the square on the left side. Keep in mind that we need to consider both the positive and negative square roots of x^2 to account for all possible solutions:
±sqrt(x^2) = ±sqrt(2^2x)
Now we have two separate equations to solve, one with the positive square root and one with the negative square root:
1) sqrt(x^2) = sqrt(2^2x)
2) -sqrt(x^2) = -sqrt(2^2x)
Simplifying each equation:
1) x = 2^x
2) -x = -2^x
Let's work with the first equation, x = 2^x, to find its solution.
To solve for x, we can use a trial and error method or use algebraic techniques. In this case, using algebraic techniques will help us find the solution more efficiently.
Let's start by graphing both sides of the equation, y = x and y = 2^x, on the same set of axes:
The graph of y = x is a straight line passing through the origin with a slope of 1.
The graph of y = 2^x is an increasing exponential curve passing through the point (0,1).
By observing the graph, we can see that there is only one point of intersection. Consequently, there is only one solution to the equation x = 2^x.
Using numerical methods, we can approximate this solution to be x ≈ 2. This means x is approximately equal to 2.
Therefore, the solution to the original equation x^2 = 4^x is x ≈ 2.