Posted by Helga on .
How do I calculate 4^n=1/(2(nn^2))? Thanks.

Algebra 
MathMate,
cross multiply to get:
2(nn^2)4^n=1
(nn^2)4^n=1/2
Since there is no more division and the righthand 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,
Is there a more elegant solution than trial and error?

Algebra 
MathMate,
You have an excellent question!
In other words, your question was more like:
Find all solutions for n such that
4^n=1/(2(nn^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^x1/(2*(xx^2))
and find the derivative
f'(x)=log(4)4^x+(12x)/(2(xx^2)^2)
and finally calculate a better approximation of x as
f1(x)=xf(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.