How to simplify these exponents?

[mathnoob2022]

According to quotient rule, if we have two exponents with the same base - we need to subtract them. I tired to simplify this fraction (photo attached below), but I get totally different answer. My answers was: 1 over Y to the negative 3.

 

[Deleted member 4993]

According to quotient rule, if we have two exponents with the same base - we need to subtract them. ......

Correct. But

y^-3/y^3 = y ^(-3 - 3) = y^(-6) = 1/y^6

When you subtract 3 from -3 → you should get (-6).

 

[BigBeachBanana]

According to quotient rule, if we have two exponents with the same base - we need to subtract them. I tired to simplify this fraction (photo attached below), but I get totally different answer. My answers was: 1 over Y to the negative 3.

To get rid of the negative exponent, [imath]y^{-3}[/imath], in the numerator, we can multiply by [imath]y^{3}[/imath]. However, we must do the same in the denominator. So we have:

\(\displaystyle \frac{y^{-3}}{y^3}\cdot \frac{y^3}{y^3}\)

Now apply the product exponential rule \(\displaystyle a^m\cdot a^n=a^{m+n}\):
\(\displaystyle \frac{y^{-3+3}}{y^{3+3}}=\frac{y^0}{y^6}=\frac{1}{y^6}\)

 

[mathnoob2022]

Correct. But

y^-3/y^3 = y ^(-3 - 3) = y^(-6) = 1/y^6

When you subtract 3 from -3 → you should get (-6).

I still struggle with this problem I've redone it and got a different answer:

 

[Dr.Peterson]

I still struggle with this problem I've redone it and got a different answer:

If you do it this way, then you need to realize that division means multiplying by the reciprocal of the divisor. The divisor here is y³, whose reciprocal is 1/y³. So you need to multiply 1/y³ * 1/y³ = 1/y⁶.

You didn't take the reciprocal.

 

[anonymoushehe]

I still struggle with this problem I've redone it and got a different answer:

An easy way is to just bring y^-3 down to the denominator which automatically changes the exponent to a positive. When you encounter other questions where the exponent is a negative in the denominator simply bring it up to the numerator.
It's important that you always want to remove negative exponents first before further simplifying. I've provided some other examples to help your understanding.

y^-3/y³ = 1/y³ * y³ = 1/y⁶

x²/y^-2 = (x² * y²)/1 = x²y²

y^-4/y^-2 = y²/y⁴ = 1/y²

 

[JeffM]

An easy way is to just bring y^-3 down to the denominator which automatically changes the exponent to a positive. When you encounter other questions where the exponent is a negative in the denominator simply bring it up to the numerator.
It's important that you always want to remove negative exponents first before further simplifying. I've provided some other examples to help your understanding.

y^-3/y³ = 1/y³ * y³ = 1/y⁶

x²/y^-2 = (x² * y²)/1 = x²y²

y^-4/y^-2 = y²/y⁴ = 1/y²

This reply needs to be fixed. As it is, this student does not understand how to divide fractions and so makes repeated mistakes in cancellation. It does not help to misuse notation.

[math]y^{-3}/y^3 \ne 1/y^3 * y^3 = \dfrac{1}{y^3} * y^3 = \dfrac{1}{y^3} * \dfrac{y^3} = \dfrac{y^3}{y^3} = 1.[/math]
I know that the quoted helper actually meant [imath]y^{-3}/y^3 = 1/(y^3 * y^3) = 1/y^6[/imath], but the student, who is already confused, cannot see that the quoted helper forgot PEMDAS.

To the original poster, stop cancelling because you do not know how to do it correctly.

[math]\dfrac{y^{-3}}{y^3} = \dfrac{\dfrac{1}{y^3}}{\dfrac{y^3}{1}} = \dfrac{1}{y^3} *\dfrac{1}{y^3} = \dfrac{1}{y^6} = y^{-6}.[/math]
Now a quick way to do this is

[math]\dfrac{y^{-3}}{y^3} = y^{-3} * \dfrac{1}{y^3} = y^{-3} * y^{-3} = y^{-6}.[/math]

 

[Steven G]

You clearly replaced 1/y^3 with y^3/1=y^3. Reciprocal's are not always equals. Does 1/3 = 3/1??

 

[JeffM]

You clearly replaced 1/y^3 with y^3/1=y^3. Reciprocal's are not always equals. Does 1/3 = 3/1??

Sorry. I don’t see it. Must be having an off day.

 

[topsquark]

Sorry. I don’t see it. Must be having an off day.

I think he was responding to post #6, not you. If he would have quoted who he was responding to....

-Dan