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View Full Version : A question on "Powers"



John Whitaker
05-12-2006, 11:31 AM
I'm not sure whether I can put an exponent up, so I write the literal form:
"13 to the 23rd power." None of my books on Beginning Algebra tell me whether I can know the actual number of the result without doing 23 separate multiplications... or, perhaps they do, and I am too thick to pick it up. Can you help? Thank you. John

JOSE JAIME-RODRIGUEZ
05-12-2006, 11:42 AM
first, what you can do is post the whole problem where 13^23 is involved in so I can know which directions the book tells you to do

pka
05-12-2006, 11:43 AM
It depends on how much work you want ot do.
\left( {13} \right)^{23} = \left( {13} \right)\left( {169} \right)^{11} = \left( {13} \right)\left( {169} \right)\left( {28561} \right)^5 = \left( {13} \right)\left( {169} \right)\left( {28561} \right)\left( {{\rm{815730721}}} \right)^2

John Whitaker
05-12-2006, 01:00 PM
Jose,
Thank you. The original problem is: 100(9/10ths) to the 26th power.

PKA,
Thank you. I'm going to take that as a "no" to my question.

John Whitaker
05-12-2006, 01:13 PM
PKA,
On second thought, I can see where you got (169), but where did the exponent 11 come from?
John

Denis
05-12-2006, 01:16 PM
John, pka is telling you only 5 multiplications are required, not 23:
13 * 169 * 28561 * 815730721 * 815730721 : that'll give you same as 13^23
(in the future, show stuff like "13 to the 23rd power" as 13^23)

"The original problem is: 100(9/10ths) to the 26th power"
That's kinda unclear, John: is it [100(9/10)]^26 or 100 * (9/10)^26 ?

pka
05-12-2006, 01:18 PM
\L
\begin{array}{l}
\left( {13} \right)^{23} = \left( {13} \right)^{22 + 1} = \left( {13} \right)\left( {13} \right)^{22} = \left( {13} \right)\left( {13^2 } \right)^{11} = \left( {13} \right)\left( {169} \right)^{11} \\
\end{array}

John Whitaker
05-12-2006, 01:27 PM
Denis,
[100(9/10)]^26. I am new here and assume that the ^ sign indicates that what follows is an exponent.

PKA,
I think I can get it now... I think... we'll see.

John Whitaker
05-15-2006, 09:57 AM
Please explain… looking at: (2^2)(2^3) = 2^2+3 = 2^5
2x2 = 4 2x2x2 = 8 4x8=32 With regard to 2^5:
1. 2x2=4
2. 2x4=8
3. 2x8=16
4. 2x16=32 That is only 4 multiplications, not 5. ????

Would also like to know how PKA writes those big numbers.
Thank you. John

stapel
05-15-2006, 10:28 AM
Would also like to know how PKA writes those big numbers.
For explanations of using LaTeX, please follow the links in the "Forum Help" pull-down menu at the very top of the page. Thank you.

Eliz.

Denis
05-15-2006, 10:28 AM
Please explain… looking at: (2^2)(2^3) = 2^2+3 = 2^5
2x2 = 4 2x2x2 = 8 4x8=32 With regard to 2^5:
1. 2x2=4
2. 2x4=8
3. 2x8=16
4. 2x16=32 That is only 4 multiplications, not 5. ????

Would also like to know how PKA writes those big numbers.
Thank you. John

Good point, John. 2^5 does require 4 multiplications.

So the rule is: a^n requires n-1 multiplications.

I goofed here; should be 4 multiplications, of course (only 4 multiplication signs!):
"John, pka is telling you only 5 multiplications are required, not 23:
13 * 169 * 28561 * 815730721 * 815730721 : that'll give you same as 13^23"

I shudda said:
"John, pka is telling you only 4 multiplications are required, not 22"

John Whitaker
05-15-2006, 11:04 AM
I think I have it... 2^5 = 2x2x2x2x2 = 2x2=4 x2=8 x2=16 x2=32
I was forgetting to count the first 2 as the first power. Only 4 multiplications are required, not 5. If this is correct, please confirm. Thank you. John

Denis
05-15-2006, 01:49 PM
CORRECT! Did you not see my previous post?

Anyway, John, what is the "purpose" of your initial question?
You mention 13^23; then mention your calculator can't handle that;
we need to go 13*13*13....(* is multiplication sign; 22 multiplications);
pka then gave you a sample of a possible short cut...

Short cuts are really only "tricks" to go faster; for instance:
by hand, 99 * 87 = ?
You can do a long multiplication to get 8613;
or you can use a trick and do it in your head, like:
87 * 100 = 8700
8700 - 87 = 8613

Or if you want 13^3 and "know" that 13^2 = 169,
then only one multiplication: 169 * 13.
...but then, the 169 required a multiplication even if you "knew" it; so I guess you can't win :shock:

John Whitaker
05-15-2006, 02:57 PM
Denis, thank you.
Q: "John, what is the "purpose" of your initial question?"
A: I was searching for the shortcut (if there was one) to the task of determining the literal solution to 13^23 without going through 22 separate multiplications.
I am still studying PKAs feedback and need to give that more time before asking for further help. To understand his responses, I feel I need to know more than i do... so I'm into several books as we speak.
Thanks.
John

Denis
05-15-2006, 04:41 PM
But WHY, John?
pka told you it depends on how much work you want to do!

What he's showing is: 13^1 * 13^2 * 13^4 * 13^8 * 13^8

Notice sum of powers = 23

This arrangement does same thing: 13^7 * 13^16 : only 1 multiplication.

Or 13^4 * 13^9 * 13^10
Or 13^6 * 13^6 * 13^6 * 13^5
...and a whola lotta others

John Whitaker
05-16-2006, 10:59 AM
Thank you Denis,
It's beginning to sink in. I will work on it more before I post again.
John