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n*log(n,2) = 18*10^7

 

solve for n.

log(n,2) is equivalent to log(n) with base = 2. 

Guest Jun 2, 2017
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6+0 Answers

 #1
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0

This is how you should write it:

n*log(2, n) = 18*10^7

n≈ 7.8582779588575599198532262×10^6 [Still Newton-Raphson method-See your last post]

Guest Jun 2, 2017
edited by Guest  Jun 2, 2017
 #2
avatar+26005 
+2

You could also do this by simple iteration.

 

\(\text{First scale the problem to deal with more manageable numbers. Let }x=\frac{n}{18\times10^7}\\.\\\text{Then the equation becomes: }x\log_2(18\times10^7x)=1 \\.\\\text{Rewrite in the form: }x=\frac{1}{\log_2(18\times10^7x)} \text{ and use a simple iterative approach.}\\.\\ x_0=1\\ x_1=\frac{1}{\log_2(18\times10^7*x_0)}\rightarrow 0.036465\\ x_2=\frac{1}{\log_2(18\times10^7*x_1)}\rightarrow0.044158\\ ...\\ x_8=\frac{1}{\log_2(18\times10^7*x_7)}\rightarrow0.0436571\\.\\ n=0.0436571\times18\times10^7\approx7858278\)

Alan  Jun 3, 2017
edited by Alan  Jun 3, 2017
 #3
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+1

Thanks very much Alan. Are you familiar with "Lambert W function?", sometimes called "Product log function?". Thank you again.

Guest Jun 3, 2017
 #4
avatar+26005 
+2

Yes, I know about the LambertW function.  This is usually written as w*e^w = x (where w is to be found given a value of x), or as w + ln(w) = ln(x).  I didn't think to seek a solution to your problem in terms of LambertW as I suspect most people using this site are not familiar with it. (Also, not immediately obvious how to manipulate your equation to LambertW form!)

.

Alan  Jun 3, 2017
 #6
avatar+26005 
+2

Thinking about this further:

 

Write equation as: n*ln(n) = 18*10^7*ln(2)

 

Let w = ln(n), so n = e^w, then we have w*e^w = 18*10^7*ln(2) so w = LambertW(18*10^7*ln(2))

 

and n = e^LambertW(18*10^7*ln(2))

.

Alan  Jun 3, 2017
 #5
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+1

Great. Thanks a lot again.

Guest Jun 3, 2017

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