MHB Exponents - Numbers with Variables

[cgr4]

(9x)^-1/2

So, I'm not entirely sure how to go about this question.
It's got a negative exponent, so I assume its 1 / something.
My guess for the answer would be:

1 / 3x

[(25xy)^3/2] / x²y

For this question, would I compute 25^3/2 and then x^3/2 y3/2 and then divide by x²y

 

[Reckoner]

(9x)^-1/2

So, I'm not entirely sure how to go about this question.
It's got a negative exponent, so I assume its 1 / something.
My guess for the answer would be:

1 / 3x

Close. Remember that in addition to taking the square root of the $9,$ you also need to take the square root of the $x.$

\begin{align*}
(9x)^{-1/2} &= \frac1{(9x)^{1/2}}\\
&= \frac1{9^{1/2}x^{1/2}}\\
&= \frac1{3\sqrt x}.
\end{align*}

[(25xy)^3/2] / x²y

For this question, would I compute 25^3/2 and then x^3/2 y3/2 and then divide by x²y

Yes. What do you get after you do that?

 

[cgr4]

For the second question. It would look something like this?

125x^3/2y^3/2

---

x²y

then I would subtract x² and y

which would equal 125x^1/2y ?

 

[MarkFL]

No, you want to apply the property of exponents:

$$\frac{a^b}{a^c}=a^{b-c}$$

In other words, for each like base, you want to subtract the exponent in the denominator from the exponent in the numerator.

 

[cgr4]

Still a little confused about the 2nd problem.

So

125x^(3/2-2)y^(3/2-2)

correct?

 

[cgr4]

resulting in 125x^-1y

125y

---

x
125y over x

 

[earboth]

Still a little confused about the 2nd problem.

So

125x^(3/2-2)y^(3/2-2)

correct?

Very close!

You have made a small typo.

125x^(3/2-2)y^{(3/2-2) should read:} 125x^(3/2-2)y^(3/2-1)

The final result should be:

$$125 \cdot x^{-\frac12} \cdot y^{\frac12}$$

Re-write this term without the fractional exponents.