Question

solve \[n\cdot \log_2 =10^9\] for n ?

Answer 1

log2 of what?

Answer 2

my bad \[ n \cdot \log_2 n =10^9\]

Answer 3

No problem. So that gives us log2n^n=10^9.

Answer 4

So far, the clues we have about n is that it needs to be a multiple of 2.

Answer 5

right, and also you can right it \[ \frac{ln (n^n)}{ln (2)}= 10^9\]
\[ ln (n^n)= 10^9 \cdot ln (2)\]

Answer 6

write*

Answer 7

how to solve the LHS?

Answer 8

Let's form an equation to solve this, shall we?
\[\frac {2^{n}}{2} \times n = 10^{9}\]

Answer 9

I am not sure, \[n\cdot log_2(n)\]=\[e^{(\frac{n*ln(n)}{ln(2)})}\]=\[n^{(\frac{n}{ln(2)})}\]

Answer 10

@ganeshie8

Answer 11

@Hero , @iambatman

Answer 12

@zepdrix "NOON"

Answer 13

@swaschan \[\frac{n\cdot2^n}{2} \neq n\cdot log_2(n)\]

Answer 14

@mathstudent55

Answer 15

@SithsAndGiggles

Answer 16

@dumbcow

Answer 17

there are no basic ways to solve this. We need something called the product log,

http://www.wolframalpha.com/input/?i=x*log_2%28x%29%3D10%5E9

Answer 18

@wio how do these work?

Answer 19

Raise both sides to the power of \(2\), and:\[
n^n=2^{10^9} = (1024)^{10^8}
\]We want the base and exponent to have the same number of digits.

Answer 20

Raise both sides to the power of \(2\), and:\[
n^n=2^{10^9} = (1024)^{10^8}
\]We want the base and exponent to have the same number of digits.
Base has \(4\), while exponent has \(8\).
If we divide exponent in half, which would square the base, we would have the base having \(8\) digits and the exponent having \(7\).

So \(n\approx 1024^2\) is a good start. You would want to use newton's method to get more accurate answers. Either way it seems like \(10^7<n<10^8\).

Answer 21

2^(10^9) =/= (2^10)^9

Answer 22

n^ n = 2^(10^9)
n^n = 2^ 1000000000

Answer 23

n = e^(W(1000000000 log(2))) ~ 3.96201*10^7

Answer 24

W(z) is the product log function

Answer 25

\[A_{n \times 1}, B_{n \times n} , C_{n \times 1}\] are three matrix, proof that \[A^TBC=\sum_{i=1}^{n}B_{ij}[AC^T]_{ji}\]