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
slove : x^2=4^x
2 answers
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.
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.
2 answers