# Math Help - solving an inequality with exponents and polynomials

1. ## solving an inequality with exponents and polynomials

i need to solve for n.

1 / (n+1)(2^(n+1)) < .0001

(n+1)(2^(n+1)) > 10,000
(n+1)(2(2^n)) > 10,000
(n+1)(2^n) > 5,000

from there i am stuck. i tried using logarithms and it didn't help either. is there any way to solve this by hand or do i need to use an equation solver program for this?

2. Hello oblixps
Originally Posted by oblixps
i need to solve for n.

1 / (n+1)(2^(n+1)) < .0001

(n+1)(2^(n+1)) > 10,000
(n+1)(2(2^n)) > 10,000
(n+1)(2^n) > 5,000

from there i am stuck. i tried using logarithms and it didn't help either. is there any way to solve this by hand or do i need to use an equation solver program for this?
Your working is fine, but you certainly won't be able to solve this inequality algebraically, so some sort of numerical method is inevitable.

One thing, though. Since the variable is $n$ and not $x$, perhaps it's an integer that you're looking for. In which case it doesn't take a moment. $2^9 = 512$. So $10\times 2^9 = 5120$. So $9$ is the first integer value of $n$ for which $(n+1)2^n > 5000$.

Is that any help?

3. Originally Posted by oblixps
i need to solve for n.

1 / (n+1)(2^(n+1)) < .0001

(n+1)(2^(n+1)) > 10,000
(n+1)(2(2^n)) > 10,000
(n+1)(2^n) > 5,000

from there i am stuck. i tried using logarithms and it didn't help either. is there any way to solve this by hand or do i need to use an equation solver program for this?
1. If $n \in \mathbb{N}$ then you can plug in some numbers to find n = 9 will satisfy this inequality.

2. If $n \in \mathbb{R}$ then you can use Newton's method to get an approximative value of n: Since $(n+1) \cdot 2^n$ is incresing for all n > 0 you can solve the equation for n:

$(n+1) \cdot 2^n - 5000=0~\implies~ n \cdot 2^n + 2^n - 5000=0$

3. Newton's method:

$n_0 = 5$

$n_{k+1}= n_k - \dfrac{n_k \cdot 2^{n_k} + 2^{n_k} - 5000}{2^{n_k}(1+(n+1) \cdot \ln(2))}$

(Don't forget to use the product rule)

Remark: i) Use the ANS-key of your calculator to get the final result. (n = 8.970103850...)
ii) It took me mor than 12 steps to reach this result which is very unusual with this method.