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John Whitaker
05-13-2006, 07:02 PM
Continuing from yesterday's "A Question on Powers (http://www.freemathhelp.com/forum/viewtopic.php?t=14540)"

PKA & Denis were kind enough to help with this, and I thought I had it; but I was wrong. I should have used a smaller exponent that my calculator can handle. I could not follow:

(13)^23 = (13)^22+1 = (13)(13)^22 = (13)(13^2)^11 = (13)(169)^11...

Also puzzling was:

(13)^23 = (13)(169)^11 = (13)(169)(28561)^5 = (13)(169)(28561)(815730721)^2

started to make sense to me, but I lost it. I figured the first exponent "11" came from (13)(13^2), but I can't figure how the exponent "5" comes into play.

I really need to know this and appreciate the patience that has been shown to me.
John

pka
05-13-2006, 08:02 PM
LEARN THE BASIC RULES!
\L
\begin{array}{l}
\left( {x^a } \right)^b = \left( {x^b } \right)^a = x^{ab} \quad \Rightarrow \quad \left( {13^2 } \right)^{11} = \left( {13^{11} } \right)^2 = 13^{22} \\
\left( {x^n } \right)\left( {x^m } \right) = x^{n + m} \quad \Rightarrow \quad \left( {13^{11} } \right)\left( {13^{12} } \right) = 13^{23} \\
\left( {x^{2n} } \right)\left( {x^1 } \right) = x^{2n + 1} \quad \Rightarrow \quad \left( {13^{12} } \right)\left( {13^1 } \right) = \left( {13^2 } \right)^6 \left( {13^1 } \right) \\
\quad \\
9^{43} = \left( {3^2 } \right)^{43} = 3^{86} \\
9^{43} = 9^{40 + 3} = \left( {9^{40} } \right)\left( {9^3 } \right) = \left( {3^{80} } \right)\left( {3^6 } \right) = 3^{86} \\
\end{array}

stapel
05-13-2006, 10:21 PM
Continuing from yesterday's "A Question on Powers (http://www.freemathhelp.com/forum/viewtopic.php?t=14540)"
Please post replies to the originating thread (http://www.freemathhelp.com/forum/viewtopic.php?t=14540). Thank you.

Eliz.