arithematic problem

[ms.cupcake]

arithematic problem

6^3 x 6^4 x 6^5 x....6^(p-1) x 6^p = (1/6)^-900. what is p? help me please. im stuck since (1/6)^-900 is at infinity. thanks for advance.

 

[pka]

6^3 x 6^4 x 6^5 x....6^(p-1) x 6^p = (1/6)^-900. what is p? help me please. im stuck since (1/6)^-900 is at infinity. thanks for advance.

Because we add exponents you want to solve \(\displaystyle \sum\limits_{k = 3}^P k = 900\).

 

[JeffM]

\(\displaystyle \left(\dfrac{1}{6}\right)^{-900} = \dfrac{1}{\left(\dfrac{1}{6}\right)^{900}} = \dfrac{1}{\dfrac{1^{900}}{6^{900}}} = \dfrac{1}{1} * \dfrac{6^{900}}{1^{900}} = \dfrac{6^{900}}{1} = 6^{900} \ne \infty.\).
.
Why in the world would you think that it is infinity: it is just a relatively large number?

 

[ms.cupcake]

Aha you notice that too. Btw,thanks for your help. im confuse which site to post so i post on both.

 

[lookagain]

Just a note:
none of my business what another helper does (ain't owner or moderator);

I disagree, Denis. You don't need to be the owner or a moderator to critique me on this site.
I expect everyone here to think what I type is also some of their business, whether they bother to type
anything in response to me in a thread or not. And as opposed to what JeffM stated for himself months ago,
I will never think that I will post as I see fit, because I know of many poor judgments I have.