A question on "Powers"

John Whitaker

Junior Member
Joined
May 9, 2006
Messages
89
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
 
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.
\(\displaystyle \left( {13} \right)^{23} = \left( {13} \right)\left( {169} \right)^{11} = \left( {13} \right)\left( {169} \right)\left( {28561} \right)^5 = \left( {13} \right)\left( {169} \right)\left( {28561} \right)\left( {{\rm{815730721}}} \right)^2\)
 
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.
 
On second thought

PKA,
On second thought, I can see where you got (169), but where did the exponent 11 come from?
John
 
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 ?
 
\(\displaystyle \L
\begin{array}{l}
\left( {13} \right)^{23} = \left( {13} \right)^{22 + 1} = \left( {13} \right)\left( {13} \right)^{22} = \left( {13} \right)\left( {13^2 } \right)^{11} = \left( {13} \right)\left( {169} \right)^{11} \\
\end{array}\)
 
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.
 
A Question On "Powers"

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
 
Re: A Question On "Powers"

John Whitaker said:
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.
 
Re: A Question On "Powers"

John Whitaker said:
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"
 
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
 
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:
 
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
 
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
 
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