Question

If (a^m)*(a^3)=(1/a^2), what is a? Could someone give me an explanation as well?

Answer 1

Are you sure you are asked for a, not m?

Answer 2

Unfortunately, they asked for a. If it was m, then I already know what it would be

Answer 3

Then a could be any nonzero number.

Answer 4

I see. Is there a rule that gives that away? If it was asking for m, then the exponent would be -5 right? I think there was a typo on my homework since it doesn't make much sense to ask for a.

Answer 5

Yes, the exponent is -5.

Answer 6

The way you know it is -5 is by using the rule of exponents which is only true if a is not zero.

Answer 7

Simply because division by 0 is undefined.

Answer 8

Alright I'm getting it now. Then a can be any number such as 2,3,4 and still be 1/a^2 then?

Answer 9

Correct.

Answer 10

Okay, thank v much for the explanation

Answer 11

Would you like an explanation why division by 0 is undefined?

Answer 12

No one really elaborated on it to me for. So yes, thank you, I would like to know actually.

Answer 13

what grade level?

Answer 14

Pre-calculus now

Answer 15

let me explain it using the exponent rule:

a^m = a*a*a*...*a

right?

Answer 16

Yes, that is correct.

Answer 17

That is for m = any positive integer: 1, 2, 3, 4, ...

right?

Answer 18

What about for the negative integers: -1, -2, -3, -4,...

what is the rule?

Answer 19

$$\Huge a^{-m}=?$$

Answer 20

Then the produce would have to become a fraction when a negative integer is the exponent, correct?

Answer 21

product*

Answer 22

$$\Huge a^{-m}=\dfrac{1}{a^m}$$

Answer 23

Yes, just like that.

Answer 24

Now, we have a rule for positive integers and negative integers, what about a 0 exponent?

Answer 25

I'm curious though, if a or m was equal to 1, then wouldn't that change the product too?

Answer 26

If 0 is the exponent then the problem would equal to 1

Answer 27

If a or m was 1 you substitute 1 into the above formula and it will work out fine :-)

Answer 28

I understand much better already. Thanks once again^^

Answer 29

The real question becomes WHY is $$\Huge a^{0}=?=1$$for any nonzero a

Answer 30

Is it not because the base is being divided by itself I wonder? If it were to have the exponent of 1 then the base would stay the same.

Answer 31

What you are wondering is EXACTLY correct :D

Answer 32

Jeez, I think I actually do not really understand how any number can become 1 just from the exponent of 0 though since it's more based on roots and and being multiplied repeatedly and no division. Is there a clearer way to explain this?

Answer 33

Unless division is really behind it

Answer 34

$$\Huge a^{0}=\dfrac{a}{a}$$

Answer 35

Yep never mind, I see very clearly now

Answer 36

Now, we can try to understand why this is true for EVERY real number except 0...ready?

Answer 37

Sure, visuals help very much^^

Answer 38

"never mind"?

Ok, you explain why it doesn't work for 0 to me

:-)

Answer 39

0 wouldn't work in place of a(the base) since 0 to any power at all would remain zero either way. Then it would be just 0 divided by 0

Answer 40

Which is undefined?

Answer 41

Am I on the wrong track?

Answer 42

Yes, you are the right track:

$$\Huge 0^{0}=\dfrac{0}{0}$$

Answer 43

for ALL the other numbers that^ equals 1, right?

Answer 44

Yes.

Answer 45

But, it equals 1 because $$\Huge\dfrac{a}{a}= a\cdot\cfrac{1}{a}$$

Answer 46

$$\Huge\dfrac{a}{a}= a\cdot\cfrac{1}{a}=1$$

Answer 47

Yes, since the a's cancel out in the center to equal 1

Answer 48

Which zero could never achieve if it was a

Answer 49

Try not to think in terms of "canceling" yet.

Just take any number and multiply it by the the result of 1 divided by the same number.

Like 5/5 = 5 * (1/5) = 5 * (0.2) = 1

Answer 50

Oh I see. I'm sorry. Then in the end we are dividing the same number together to equal And with your example of 5 * (0.2)=1 then certain decimals work perfectly fine to make 1, but it won't come up in exponential problems will it?

Answer 51

Yes, it does come up in the exponential problem because $$\Huge\dfrac{a}{a}= a\cdot\cfrac{1}{a}=1$$
$$\Huge\dfrac{0}{0}= 0\cdot\cfrac{1}{0}=0 $$ because 0 times any real number is 0

Answer 52

Ah, I get it. (0.2) is also (1/5) which is usable for a power.

Answer 53

Yes, but 1/0 is not "usable"

Answer 54

Alright, any integer is doable besides 0 since most circumstances that regular integers go through wouldn't work, such as being divided by other numbers. I'm sorry, it must be a pain for you to write out these equations. Am I miss anything else important?

Answer 55

Dividing by 0 would mean multiplying by 1/0

but 0 has no reciprocal because

0 times any number is 0, not 1.

Answer 56

Therefore, division by 0 has no meaning in the set of real numbers.

Answer 57

Okay, it becomes clearer now that I also keep in mind that though 0 is real number and an integer it doesn't have a reciproval. I think what you showed me was the inverse property of multiplication right?

Answer 58

Oh dear, I kept you here for an hour. :S

Answer 59

Good job :D

Answer 60

Thank you, I really appreciate you writing these examples and taking your time out to help me :)

Answer 61

$$\Huge a^0=\dfrac{a}{a}= a\cdot\cfrac{1}{a}=1$$

$$\Huge 0^0=\dfrac{0}{0}= 0\cdot\cfrac{1}{0}=0$$

Answer 62

See^ the contradiction?

Answer 63

Yes, 0 doesn't follow the same rules

Answer 64

Because it has its OWN rule that says 0*(any number) = 0

Answer 65

So, 1/0 cannot be (any number)

Answer 66

Then a good summary would be that 0^0 or to any power would mostly be meaningless?

Answer 67

Since 0/0=0 and 0*0=0 always

Answer 68

Well 0/0 is considered undefined, but with 1/0 does it =0 or is undefined?

Answer 69

The expression 0^0 has no meaning.

Because division by 0 is undefined.

Answer 70

1/0 is undefined too.

Answer 71

Okay, I think I fully understand by now.