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

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 ?

\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

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