Need Maths Help????

Question

For all n>=4,prove that 2^n>=4n.
[Not in graph..if possible any other way]

Answer 1

You could try a formal proof by induction which is kinda cool. You have to show 2 things. See below

First we need to rewrite the equation so that n is always greater or equal to 4. let n = m+3. The induction is now done on m.

Thing 1 show it is true for m = 1 (or n=4)
LS: 2^n = 2^(1+3) = 16, and
RS: 4n = 4(1+3) = 16
so the statement is true at m=1 (n=4)

Thing 2: Make believe it is true for a value m and show it it still true for a value m+1
So: 2^(m+3) ≥ 4(m+3) is a true statement
What about m+1?
2^(m+3+1) ? 4(m+3+1)
LS: 2^(m+3+1) = 2*2^(m+3)
RS: 4(m+3+1) = 4(m+3) + 4

to show LS≥RS first divide both sides by 2
2^(m+3) ? 2(m+3) + 2
but 2^(m+3) ≥ 4(m+3)
and 4(m+3)≥ 2(m + 3) + 2 because (4m + 12) > (2m + 8)

So the statement is true for m + 1

and that completes the proof because if the statement is true for m = 1 and if it is true for m then it is true for m+1 then it is true for all m. n = m+3 it is true for all n≥4
QED

Answer 2

Clearly it's true for n=4
2^4 = 16 ≥ 4*4 = 16
Now, if it's true for some n, then
2^n ≥ 4n then
2^(n+1) = 2*2^(n) = 2^n + 2^n while
4(n +1) = 4n + 4 therefore
2^n + 2^n ≥ 4n+4 and (since we have 2^n ≥ 4n) it follows that
2^(n+1) ≥ 4n for all n ≥ 4

HTH

Doug

Answer 3

2^n >= 4n , n=>4
log 2^n >= log 2^2 + log n
nlog2 >= 2log2 +logn
nlog2 -2log 2 >= logn
(n-2)log2 >= logn
n must be greater than 0, else logn is undefined.
if n = 4 we have 2log2 >=log4 which is true
If n=5 we have 3log2 => log5 which is also true since 3log 2 = log 2^3 = log 8 which is greater than log 5.
It is obvious that log2^(n-2) =>log n for all n >4

Answer 4

Think of it a different equation. If 2^n is always greater, then the difference between them must be greater than zero.

2^n-4n must be positive. This equation is zero at n=4 and approaches infinity as n approaches infinity.