Question

I have a bunch of problems like this and I need help understanding how to do them (5)^-5(5)^7
My choices are
A. 1/25
B. 25
C. 12
D. 2

Answer 1

\[5^{-5}*5^7\]

Answer 2

First, do you understand what the notation means?

Answer 3

we'll for the sake of my confusion go with no just to get an explanation

Answer 4

Okay. Think of the exponent (the little number up top) as a shorthand.

\[x^2 = x*x\]\[x^3=x*x*x\]\[x^4=x*x*x*x\]\[x^{537962} = x*x*x*x*\text{just kidding!}\]
Okay so far?

Answer 5

so far so good

Answer 6

Now, if we have a negative exponent, that's the same as:

\[x^{-2} = \frac{1}{x^2} = \frac{1}{x*x}\]\[x^{-3} = \frac{1}{x^3} = \frac{1}{x*x*x}\]and in general,
\[x^{-n} = \frac{1}{x^n}\]and\[\frac{1}{x^{-n}} = x^{n}\]

Answer 7

ok I've got you so far.

Answer 8

Now, the property of exponents this problem is apparently to exercise is this one:

\[x^{n}*x^{m} = x^{n+m}\]

In other words, when you multiply exponents \(\bf\text{and they have the same base}\) you add the exponents to get the exponent of the product.

Examples:
\[2^2*2^3 = (2*2)*(2*2*2) = 2*2*2*2*2 = 2^{2+3} = 2^5\]Checking:\[2^2=4, 2^3 = 8, 4*8 = 32, 2^5 = 2*2*2*2*2= 32\]

\[2^{-2}*2^4 = \frac{1}{2^2}*2^4 = \frac{1}{2*2}*(2*2*2*2) = 2*2 = 4\]\[2^{-2}*2^2 = 2^{-2+4} = 2^2 = 4\]

Answer 9

sorry, I mistyped the last line, should be \[2^{-2}*2^4 = 2^{-2+4} = 2^{2} = 4\]

Answer 10

two other things to remember are that \[x^1 = x\]and \[x^0=1\](the latter being true so long as \(x\ne0\), if \(x = 0\) it is undefined)

Answer 11

So your problem is \[5^{-5}*5^{7} = \]what is the answer in exponential form, after you've added the exponents?

Answer 12

5^2?

Answer 13

Yes! Now, if you convert that to a number such as found in your answer choices, which one would it be?

Answer 14

A.?

Answer 15

What does \(5^2\) mean?

Answer 16

mistype sorry I meant B.

Answer 17

That's correct. \[5^2 = 5*5 = 25\]

Answer 18

Here, I'll give you another problem with an answer from that list:

\[5^1*5^{-3} = \]

Answer 19

Thanks for the Help I understand what im doing now :)

Answer 20

erm lets 5^-2= 25?

Answer 21

Well, the first part of the answer is correct.

\[5^1*5^{-3} = 5^{1-3} = 5^{-2}\]Remember, if you have a negative exponent, put it on the other side of the division bar and make it a positive exponent:
\[5^{-2} = \frac{1}{5^2} = \]

Answer 22

oh so 1/25

Answer 23

Right!

There's only 1 number for which \[x^n = x^{-n}\]Can you guess what it might be?

Answer 24

2?

Answer 25

Hint: it appears in the question :-)

Answer 26

I give whats the answer lol?

Answer 27

It's not 2: say \(n = 3\), then
\[2^3 = 8\]but \[2^{-3} = \frac{1}{2^3} = \frac{1}{8}\]and that isn't the same as \(8\), I hope you'll agree!

How about 1?

\[1^n = 1*1*1*1*...*1=1\]
\[1^{-n} = \frac{1}{1^n} = \frac{1}{1*1*1*1*...*1} = \frac{1}{1} = 1\]

Answer 28

so 1 I take it lol

Answer 29

Yes. 1 is the only number that works. A gentle way of suggesting that if the answer to
one problem is \(x^n =\text{some number}\) then you better not get \(x^{-n} = \text{same number}\) unless \(x = 1\)...

Answer 30

Well thank you whpalmer I have repaid you with the medals and a fan :)

Answer 31

Hopefully that is enough to get you through the current homework with flying colors :-)