The equation for the pattern 2, 4, 16, 256, 65536 is established as y = 2^(2^x). This equation accurately reflects the sequence when x starts at 0, yielding the correct outputs for each corresponding value of x. The initial misunderstanding regarding the starting point of x was clarified, emphasizing that x=0 must be used to achieve the desired results. The discussion highlights the importance of correctly identifying the base case in exponential functions.
PREREQUISITESMathematicians, programmers, and students working on projects involving exponential functions and sequences will benefit from this discussion.
But this doesn't help me much becasue it doesn't find my numbers i need in an equation...Try writing all of them as powers of 2
I had already thought of this but it does not work...Here is the pattern it gives you.f(x) = 2^2x
Acually no, it's not for homework...its for a project I am working on(programming).I don't think we should be giving explicit formulas anyway, since it sounds like homework.
Originally posted by Qwerty
Acually no, it's not for homework...its for a project I am working on(programming).
And I don't need your help anymore, I found the answer to my question myself.
Answer: 2^(2^x)
if you first put x=1 (as you requested) to this equation you get 2^(2^1)=4 which is not the first number 2 ofcourse.Originally posted by Qwerty
I have the following pattern and i need an equation that if x = 1 then y will equal the fist number of it, if x = 2, then y will equal the 2nd number.
The pattern is:
2, 4, 16, 256, 65536
Hint: It keeps squaring itself.
Any ideas?
Yes this is an error in my first post, 0 is supposed to be the first number subsituted into the equation.if you first put x=1 (as you requested) to this equation you get 2^(2^1)=4 which is not the first number 2 ofcourse.