Equations with negative powers

drunk_monk

Hi all,

I'm having problems with equations with negative powers and need a bit of help.
I'm unable to solve equations like the one below and for the last few days have been unable to find an answer (even though I'm sure it's easy).

(5xy)-2 + (4y)3
(3y)-3

If someone would be kind enough to explain this to me or point me to an online tutorial I would be very grateful,
thanks
DM

Fremen

What you posted is an expression rather than an equation. An equation has an "=" sign. Since you have both x's and y's in there, you need to specify which term you're solving for, too.

The usual text notion for exponentiation is "^", so x^y would be read "x to the power of y". For example, 2^4 = 16.

If you could clarify what it is you're trying to do, I'll try to help.

drunk_monk

thanks for the quick reply, sorry it wasn't clear. I'm trying to teach myself powers and I'm ok with the rest, it's just the negative powers are confusing me. It's just a case of simplifying it but I can't do it. I've attached a better example (I hope) of the sort of thing I'm stuck on.

Thanks

Fremen

Oh, I see what you're doing. It's called simplifying an expression rather than solving an equation.

There's just one rule and one trick you need to know. The rule is

[latex] x^{-a} = \frac{1}{x^a}[/latex]

The trick is "multiply by one in a clever way".

I'll show you what I mean by that. Anything multiplied by 1 is just that same thing, right? So suppose we want to know

[latex]\frac{x^b}{x^{-a}}[/latex]

We multiply it by one:

[latex]\frac{x^b}{x^{-a}} \times \frac{x^a}{x^a}[/latex]

which is the same as

[latex]\frac{x^b \times x^a}{1}} = x^{a+b}[/latex]

So in your case the trick is a little different. Multiply it by 1 by writing

[latex]1 = \frac{(3y)^3}{(3y)^3}[/latex]

drunk_monk

Thanks for your help fremen, it all makes sense now