Solving N(5/6)^(N-1) > 27/5 for N

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Flucky
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Hi all sorry the title couldn't be more descriptive.

I'm having a bit of a brain block with this equation (it's the very end of a problem).

N([itex]\frac{5}{6}[/itex])[itex]^{N-1}[/itex] > [itex]\frac{27}{5}[/itex]

My line of thought was to get the first N in terms of the power N-1..

How do I find N?
 
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Analytically you can't.
 
Sorry I meant how do I simplify the inequality, ie N > #
 
What dirk_mec1 said.

However, a comment in the OP raises a question: is the coefficient N the same variable as the exponent N? You seem to imply that they are different.
 
Ah I guess I've gone wrong somewhere.

Yes the N's are the same.
 
It represents the rolls of a dice. I'm going to post the full question over in the physics homework subforum.
 
Flucky said:
It represents the rolls of a dice. I'm going to post the full question over in the physics homework subforum.

Then it suffices to solve it by trial and error. An easy method is to graph the equation ##y=x\left(\frac{5}{6}\right)^{x-1}## with a calculator. That way the value(s) of ##N## that satisfy the OP will become clear. Once you found a suitable candidate for ##N##, you can rigorously prove that this ##N## is a good candidate by just plugging it in the equation and calculating manually.

There is no way to solve the equation analytically, so you will need to resort to trickery such as the above.
 
Flucky said:
Thanks but I have probably just messed up somewhere, as I don't think it will be that much hassle.

https://www.physicsforums.com/showthread.php?p=4716123#post4716123

Actually, if the problem is truly as given in the above link then the solution will involve even more hassle that what you have now. Why would you think the solution is "easy"? What do you have against plotting graphs or against using numerical methods?