How Do You Solve Equations With Negative Exponents?

[TbbZz]

Homework Statement

8x^-3 = 64

Homework Equations

None.

The Attempt at a Solution

I tried doing all sorts of things, changing 8x^-3 to (1/8x)^3 or trying to get both sides to have the same base, but couldn't get it to work.

The book I am using does not explain how to do so, I have already looked through the whole chapter that the problem is from.

Primarily, I would like assistance in understanding the concepts behind this problem. With the current problem, it is relatively easy to figure out by educated guessing & checking, but when there are more difficult numbers and more challenging problems, guessing and checking won't work. Thanks for the assistance.

 

[berkeman]

Hint: (x^-3)^3 = ?

 

[TbbZz]

Hint: (x^-3)^3 = ?

(x^-3)^3 = X^-9 but I'm not quite sure how that helps.

Thanks for the response.

 

[Dick]

Berkeman meant (x^-3)^(-1/3).

 

[TbbZz]

Got it. I didn't realize you could raise both sides to a power ((-1/3) in this case). Thanks for the help.

 

[berkeman]

Berkeman meant (x^-3)^(-1/3).

Whoops. Thanks Dick.

 

[JonF]

Have you tried dividing both sides by 8?

Then did you try expressing x^-3 as an expression with a positive exponent using the rule a^-b = 1/a^b

 

[CompuChip]

You can always raise both sides to the same power.
In fact this is exactly what you're doing when solving something like
3 x^2 = 27
If you divide out the 3 you get
x^2 = 9
and you would take the square root to get x = 3 (or - 3 of course). But you can also see it as raising both sides to the power 1/2:
(x^2)^{1/2} = x^{2 \times 1/2} = x \quad = \quad 9^{1/2} = \sqrt{9}
where the last equality is just a change of notation, so you see that \sqrt{x} = x^{1/2}.

But the raising-both-sides-to-a-power-trick works even in the case of negative and fractional exponents.