Solve an Exponent Question with Help from Experts

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Homework Help Overview

The problem involves exponential equations where the original poster seeks to determine the value of x^(y-t) given that x^y = 2 and x^t = 16. The context is centered around understanding exponent rules and notation clarity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the notation used in the problem, questioning whether it is interpreted correctly. There are attempts to clarify the expression x^(y-t) and its relation to the properties of exponents. Some participants suggest breaking down the expression into a product of exponents.

Discussion Status

The discussion is active with participants exploring different interpretations of the notation and the properties of exponents. Some guidance is offered regarding the manipulation of the expressions, but there is no explicit consensus on the interpretation of the notation.

Contextual Notes

There is a mention of the need to find values for x, y, and t, and a discussion about the constraints on x being a positive number. The original poster's uncertainty about where to start is noted, along with the varying interpretations of the notation used in the problem.

benr2424
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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:
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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!
 

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