In the equation:
Lambda1=(n+1)Lambda2
What is the trick 2 get the solution:
n=Lambda2/(Lambda1-Lambda2)
I understand Your reply fully, thanks:-)
But since there is an n on both sides the correct equation:
nLambda 1=(n+1)Lambda 2
with solution:
n=lambda 2/(lavmbda 1-lambda 2) has a different technique which I cant figure out. If You know how I would be very grateful
Do you realize how difficult it is to read your postings?
Why not learn to post in symbols? You can use LaTeX tags
[tex] n\lambda_1=(n+1)\lambda_2[/tex] gives $\displaystyle n\lambda_1=(n+1)\lambda_2 $.
It's pretty much basic algebra- you have to get n by itself so first you have to "free" it from the right side by using the "distributive" property: $\displaystyle (n+1)\lambda_2= n\lambda_2+ \lambda_2$. At that point you have
$\displaystyle n= n\lambda2+ \lambda2$
Still trying to get n by itself, subtract $\displaystyle n\lambda$ from both sides:
$\displaystyle n\lambda_1- n\lambda_2= \lambda_2$
Now use the "distributive" property again:
$\displaystyle n(\lambda_1- \lambda_2)= \lambda_2$
and, finally, divide both sides by $\displaystyle (\lambda_1- \lambda_2)$.
That's essentially just what Plato did before.