Thursday

March 5, 2015

March 5, 2015

Posted by **Helga** on Thursday, March 1, 2012 at 2:53am.

- Algebra -
**MathMate**, Thursday, March 1, 2012 at 9:18amcross multiply to get:

2(n-n^2)4^n=1

(n-n^2)4^n=1/2

Since there is no more division and the right-hand side is less than 1, we can start searching for n where n<1, say 1/2.

Try n=1/2 and see if the equation is satisfied.

Hint: one of the laws of exponents tells us that

X^(1/2)=√(X).

- Algebra -
**Helga**, Thursday, March 1, 2012 at 10:49amIs there a more elegant solution than trial and error?

- Algebra -
**MathMate**, Thursday, March 1, 2012 at 2:37pmYou have an excellent question!

In other words, your question was more like:

Find all solutions for n such that

4^n=1/(2(n-n^2))

I do not see an explicit solution to the equation. Perhaps someone else can find one.

An explicit solution is expressed as

n=expression where expression does not contain n.

Lack of an explicit solution, I proceed as follows:

1. first bound the solution.

We can conclude that for n outside of [-1,1], we cannot have the right hand side equal to -1.

2. Find approximate solutions by graphing or otherwise. By graphing, there are two solutions, at 0.5 and 0.8, approximately.

3. proceed to refine the solutions by iterations (a glamorous name for trial and error).

Here are the details of iteration using Newton's method:

For the case of n=0.5, it is exact and so requires no further iteration.

For the case of n=0.8, we can refine the solution by setting up Newton's iteration equation:

let f(x)=4^x-1/(2*(x-x^2))

and find the derivative

f'(x)=log(4)4^x+(1-2x)/(2(x-x^2)^2)

and finally calculate a better approximation of x as

f1(x)=x-f(x)/f'(x)

and proceed to calculate

x0=0.8,

x1=f1(x0)=0.787551458195669

x2=f1(x1)=0.786114835166431

x3=f1(x2)=0.786097645437988

x4=f1(x3)=0.786097643010236

x5=f1(x4)=0.786097643010236

which means that 0.786097643010236 is our (approximate) solution.

Hope I have answered your question.

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