Exponents

mickeybell

New member
Joined
Feb 12, 2014
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2
okay so my math teacher just taught us exponents today but with negative numbers. i must be really dumb not understanding or it is just plan hard. anyone can help.:confused::confused::confused::confused:
 
also it had letters like;
x-8
3x-5
also like,
3x0
------
y-3
:twisted::twisted::twisted::twisted::twisted::twisted::confused::confused::confused::confused::confused::(:(:(:(:-?:-?:idea::idea::oops::oops::oops::cry::cry::cry::cry:
 
okay so my math teacher just taught us exponents today but with negative numbers. i must be really dumb not understanding or it is just plan hard. anyone can help.:confused::confused::confused::confused:
You are not dumb, nor is it hard. It's just new.

\(\displaystyle a\ is\ a\ real\ number \ne 0 \implies a^0 \equiv 1.\) That is a definition.

\(\displaystyle a\ is\ a\ real\ number \ne 0\ and\ n\ is\ an\ integer > 0 \implies a^n \equiv a * a^{(n - 1)}.\) That too is a definition.

\(\displaystyle a\ is\ a\ real\ number \ne 0\ and\ n\ is\ an\ integer > 0 \implies a^{-n} \equiv \dfrac{1}{a^n}.\) That too is a definition.

Those three definitions completely DEFINE what we mean by an exponent that is an integer for a number that is not zero.

So \(\displaystyle 3^1 = 3 * 3^0 = 3 * 1 = 3.\)

\(\displaystyle 3^2 = 3 * 3^1 = 3 * 3 = 9.\)

\(\displaystyle 3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}.\)

That's it. It's one of those things that is so simple that you say, "Oh, there must be something I am missing here because how can they make a fuss about that? It must be super hard and I just can't get it." Nope. It really is that simple.

Now give those problems a shot, post the answers, and let us see how you did.

PS Be careful with \(\displaystyle (3x)^{-5}\ versus\ 3x^{(-5)}.\) They have different answers. Do you see why?
 
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