anonymous
  • anonymous
If (a^m)*(a^3)=(1/a^2), what is a? Could someone give me an explanation as well?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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skullpatrol
  • skullpatrol
Are you sure you are asked for a, not m?
anonymous
  • anonymous
Unfortunately, they asked for a. If it was m, then I already know what it would be
skullpatrol
  • skullpatrol
Then a could be any nonzero number.

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anonymous
  • anonymous
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.
skullpatrol
  • skullpatrol
Yes, the exponent is -5.
skullpatrol
  • skullpatrol
The way you know it is -5 is by using the rule of exponents which is only true if a is not zero.
skullpatrol
  • skullpatrol
Simply because division by 0 is undefined.
anonymous
  • anonymous
Alright I'm getting it now. Then a can be any number such as 2,3,4 and still be 1/a^2 then?
skullpatrol
  • skullpatrol
Correct.
anonymous
  • anonymous
Okay, thank v much for the explanation
skullpatrol
  • skullpatrol
Would you like an explanation why division by 0 is undefined?
anonymous
  • anonymous
No one really elaborated on it to me for. So yes, thank you, I would like to know actually.
skullpatrol
  • skullpatrol
what grade level?
anonymous
  • anonymous
Pre-calculus now
skullpatrol
  • skullpatrol
let me explain it using the exponent rule: a^m = a*a*a*...*a right?
anonymous
  • anonymous
Yes, that is correct.
skullpatrol
  • skullpatrol
That is for m = any positive integer: 1, 2, 3, 4, ... right?
skullpatrol
  • skullpatrol
What about for the negative integers: -1, -2, -3, -4,... what is the rule?
skullpatrol
  • skullpatrol
$$\Huge a^{-m}=?$$
anonymous
  • anonymous
Then the produce would have to become a fraction when a negative integer is the exponent, correct?
anonymous
  • anonymous
product*
skullpatrol
  • skullpatrol
$$\Huge a^{-m}=\dfrac{1}{a^m}$$
anonymous
  • anonymous
Yes, just like that.
skullpatrol
  • skullpatrol
Now, we have a rule for positive integers and negative integers, what about a 0 exponent?
anonymous
  • anonymous
I'm curious though, if a or m was equal to 1, then wouldn't that change the product too?
anonymous
  • anonymous
If 0 is the exponent then the problem would equal to 1
skullpatrol
  • skullpatrol
If a or m was 1 you substitute 1 into the above formula and it will work out fine :-)
anonymous
  • anonymous
I understand much better already. Thanks once again^^
skullpatrol
  • skullpatrol
The real question becomes WHY is $$\Huge a^{0}=?=1$$for any nonzero a
anonymous
  • anonymous
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.
skullpatrol
  • skullpatrol
What you are wondering is EXACTLY correct :D
anonymous
  • anonymous
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?
anonymous
  • anonymous
Unless division is really behind it
skullpatrol
  • skullpatrol
$$\Huge a^{0}=\dfrac{a}{a}$$
anonymous
  • anonymous
Yep never mind, I see very clearly now
skullpatrol
  • skullpatrol
Now, we can try to understand why this is true for EVERY real number except 0...ready?
anonymous
  • anonymous
Sure, visuals help very much^^
skullpatrol
  • skullpatrol
"never mind"? Ok, you explain why it doesn't work for 0 to me :-)
anonymous
  • anonymous
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
anonymous
  • anonymous
Which is undefined?
anonymous
  • anonymous
Am I on the wrong track?
skullpatrol
  • skullpatrol
Yes, you are the right track: $$\Huge 0^{0}=\dfrac{0}{0}$$
skullpatrol
  • skullpatrol
for ALL the other numbers that^ equals 1, right?
anonymous
  • anonymous
Yes.
skullpatrol
  • skullpatrol
But, it equals 1 because $$\Huge\dfrac{a}{a}= a\cdot\cfrac{1}{a}$$
skullpatrol
  • skullpatrol
$$\Huge\dfrac{a}{a}= a\cdot\cfrac{1}{a}=1$$
anonymous
  • anonymous
Yes, since the a's cancel out in the center to equal 1
anonymous
  • anonymous
Which zero could never achieve if it was a
skullpatrol
  • skullpatrol
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
anonymous
  • anonymous
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?
skullpatrol
  • skullpatrol
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
anonymous
  • anonymous
Ah, I get it. (0.2) is also (1/5) which is usable for a power.
skullpatrol
  • skullpatrol
Yes, but 1/0 is not "usable"
anonymous
  • anonymous
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?
skullpatrol
  • skullpatrol
Dividing by 0 would mean multiplying by 1/0 but 0 has no reciprocal because 0 times any number is 0, not 1.
skullpatrol
  • skullpatrol
Therefore, division by 0 has no meaning in the set of real numbers.
anonymous
  • anonymous
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?
anonymous
  • anonymous
Oh dear, I kept you here for an hour. :S
skullpatrol
  • skullpatrol
Good job :D
anonymous
  • anonymous
Thank you, I really appreciate you writing these examples and taking your time out to help me :)
skullpatrol
  • skullpatrol
$$\Huge a^0=\dfrac{a}{a}= a\cdot\cfrac{1}{a}=1$$ $$\Huge 0^0=\dfrac{0}{0}= 0\cdot\cfrac{1}{0}=0$$
skullpatrol
  • skullpatrol
See^ the contradiction?
anonymous
  • anonymous
Yes, 0 doesn't follow the same rules
skullpatrol
  • skullpatrol
Because it has its OWN rule that says 0*(any number) = 0
skullpatrol
  • skullpatrol
So, 1/0 cannot be (any number)
anonymous
  • anonymous
Then a good summary would be that 0^0 or to any power would mostly be meaningless?
anonymous
  • anonymous
Since 0/0=0 and 0*0=0 always
anonymous
  • anonymous
Well 0/0 is considered undefined, but with 1/0 does it =0 or is undefined?
skullpatrol
  • skullpatrol
The expression 0^0 has no meaning. Because division by 0 is undefined.
skullpatrol
  • skullpatrol
1/0 is undefined too.
anonymous
  • anonymous
Okay, I think I fully understand by now.

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