Why does this work?

[u/ShortCover4598]

Why does this law of exponents work? I've been having trouble with this in my self taught Algebra 1 course. A 9th grade level explanation would help, thanks.

[u/CaptainMatticus]

You know that a^b / a^c = a^(b - c), right?

And you should also know that a^0 = 1, for all non-zero values of a, right?

So rewrite 1 / a^x as a^0 / a^x, which would also be a^(0 - x), or a^(-x)

[u/ShortCover4598 → u/CaptainMatticus]

OMG I JUST HAD SUCH A NEURON ACTIVATION MOMENT THANK YOU SO MUCH!!

[u/Top_Bluejay_5323 → u/CaptainMatticus]

So you started with the result to get to the result??

The first thing you did was state that
1/a^c=a^(-c)

[u/CaptainMatticus → u/Top_Bluejay_5323]

That's not what I did at all. I showed a different way of understanding what was happening and got the result.

Rules for exponents:

(a^b)^c = a^(b * c) = (a^c)^b

a^b * a^c = a^(b + c)

a^b / a^c = a^(b - c)

Now anybody can understand how a^0 = 1. Start with a^n and divide by a^1

a^n / a^1 = a^(n - 1)

And a^(n - 1) divided by a^1 is: a^(n - 1) / a^1 = a^(n - 1 - 1) = a^(n - 2)

And a^(n - 2) divided by a^1 is: a^(n - 2) / a^1 = a^(n - 2 - 1) = a^(n - 3)

And so on. What happens when we get to a^(n - n)? Well that's a^0, which we already know is 1. We know that because x/x = 1 for all non-zero values of x. That's a basic identity in Algebra.

a^n / a^n = 1

a^2 / a^2 = 1

a^3 / a^3 = 1

And so on.

And since we know that a^b / a^c = a^(b - c), we can demonstrate that

a^(n - n) = a^0 = 1

Now that we definitively know that a^0 = 1, we can go one step further

1 / a^x

is nothing more than

a^0 / a^x

Which is

a^(0 - x)

Which is

a^(-x)

I cannot be more exhaustive with this and I won't waste any more time on you. The OP got it, so what's your critique all about?

[u/SufficientStudio1574 → u/Top_Bluejay_5323]

Sometimes you just need to say the same thing with different words.

[u/Shevek99]

It's defined that way to have consistence.

We want this formula to hold for all integers

a^m aⁿ = a^{m + n}

Now, making n = 0 we get

a^m a⁰ = a^m

So it must be

a⁰ = 1

Now, if we put n = -m

a^m a^-m = a^{m - m} = a⁰ = 1

Then a^-m must be the multiplicative inverse of a^m

a^-m = 1/a^m

[u/ShortCover4598 → u/Shevek99]

Thank youuu

[u/JeffTheNth]

Everyone points to the definition and such, but let me give you an example...

32 / 4 = 8

2⁵ / 2² = 2³

2^(5-2) = 2³

The negative exponent would be subtracted... If there isn't enough, it ends up as a divisor.
1/4 = 1/2² = 2^-2

having a good example in mind can help make that a little clearer... hope this helps!

[u/ShortCover4598 → u/JeffTheNth]

Thank you so much!

[u/Consistent-Annual268]

That's the literal definition of negative exponents. It is defined like that precisely so that it works nicely with the remaining exponent rules.

[u/goldenrod1956 → u/Consistent-Annual268]

Not a math guy but that was also my thought…that is the symbolism implies

[u/Top_Bluejay_5323]

a^(-1) (a^1/a^1)=a^(-1)a^1/a^1= a^(-1+1)/a^1

a^(-1+1)=a^0=1

Therefore a^(-1)=1/a^1

[u/Trimutius]

By definition really... but yeah when you reduce power you keep dividing by x, then at 0 you reach 1 and then when you reduce once more you go to 1/x

[u/CoachSevere5365]

It's interesting that nobody has mentioned logarithms in this thread.

The rules about adding and subtracting exponents apply to non-integers too. Before electronic calculators were common we used "log books" which contained tables of exponents.

To multiply numbers together you'd look up each number in a table to get its logarithm, add the logarithms, then do the reverse to get the result.

There was a bit of extra bookwork in that using the tables meant that numbers had to be scaled to be between 0 and 1, and the powers of ten (or whatever the base was) handled separately, but the fundamental principle doesn't change.