Solving Negative Exponents

  • Thread starter TbbZz
  • Start date
  • #1
37
0

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.
 

Answers and Replies

  • #2
berkeman
Mentor
58,024
8,082
Hint: (x^-3)^3 = ?
 
  • #3
37
0
Hint: (x^-3)^3 = ?
(x^-3)^3 = X^-9 but I'm not quite sure how that helps.

Thanks for the response.
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
Berkeman meant (x^-3)^(-1/3).
 
  • #5
37
0
Got it. I didn't realize you could raise both sides to a power ((-1/3) in this case). Thanks for the help.
 
  • #6
berkeman
Mentor
58,024
8,082
Berkeman meant (x^-3)^(-1/3).
Whoops. Thanks Dick.
 
  • #7
612
1
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
 
  • #8
CompuChip
Science Advisor
Homework Helper
4,302
47
You can always raise both sides to the same power.
In fact this is exactly what you're doing when solving something like
[tex]3 x^2 = 27[/tex]
If you divide out the 3 you get
[tex]x^2 = 9[/tex]
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:
[tex](x^2)^{1/2} = x^{2 \times 1/2} = x \quad = \quad 9^{1/2} = \sqrt{9}[/tex]
where the last equality is just a change of notation, so you see that [tex]\sqrt{x} = x^{1/2}[/tex].

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

Related Threads on Solving Negative Exponents

Replies
3
Views
7K
Replies
1
Views
3K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
12
Views
7K
  • Last Post
Replies
3
Views
2K
Replies
5
Views
1K
Replies
20
Views
4K
Replies
2
Views
16K
Top