Solve an Exponent Question with Help from Experts

  • Thread starter Thread starter benr2424
  • Start date Start date
  • Tags Tags
    Exponent
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
8 replies · 3K views
benr2424
Messages
3
Reaction score
0
Hello everyone!

The following question stumped me and I thought you may be able to help.

If x^y = 2 and x^t = 16 what does x^(y-t) equal?

I don't know where to start so any help would be appreciated.
Thanks,
Ben
 
Last edited:
Physics news on Phys.org
You should be clearer with the notation. By x^y-t, do you mean (x^y)-t or x^(y-t)? The notation looks like (x^y)-t, but I'm probably not wrong when guessing that you meant x^(y-t)?
 
Assuming you meant that you wanted to find [itex]x^{y-t}[/itex], if [itex]a^n \times a^m = a^{n+m}[/itex], can you split [itex]x^{y-t}[/itex] into the product of two exponents?
 
Sorry, the notation in the problem was given as such:

...what does x^(y-t) equal?
 
benr2424 said:
Sorry, the notation in the problem was given as such:

...what does x^(y-t) equal?

Do you know the rules regarding indices?
 
When you say x[tex]^{y-t}[/tex] it's the same as [tex]\frac{x^{y}}{x^{t}}[/tex]
First we need to find x, y and t.
So if x[tex]^{y}[/tex] = 2, and x[tex]^{t}[/tex] = 16, then x needs to be even. It also needs to be able to divide those numbers into whole integers. Now the only number that fits into those catagories is 2.
2, however, could be positive or negative. But the fact that x[tex]^{1}[/tex] is 2, means that it cannot be a negative number.

So we know x=2, y=1 and t=4 (2*2*2*2 = 16). If that is the case, then x[tex]^{y-t}[/tex] = 2[tex]^{1-4}[/tex] = 2[tex]^{-3}[/tex] = [tex]\frac{1}{2*2*2}[/tex] = [tex]\frac{1}{8}[/tex] = 0.125

OR: [tex]\frac{x^{y}}{x^{t}}[/tex] = [tex]\frac{2^{1}}{2^{4}}[/tex] = [tex]\frac{2}{16}[/tex] = [tex]\frac{1}{8}[/tex] = 0.125

Hope that answers your question.
 
?!

We do not know to know either x, t or y. x^y-t = x^y / x^t = 2 / 16 = 1/8. Voila.
 
Good point... Ahh well, either gives you the right answer. However, you could use it as reference for checking.
 
Thanks everyone! 1/8 is the correct answer!