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Exponents - Numbers with Variables

cgr4

New member
Aug 31, 2013
7
(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 253/2 and then x3/2 y3/2 and then divide by x2y
 

Reckoner

Member
Jun 16, 2012
45
(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 253/2 and then x3/2 y3/2 and then divide by x2y
Yes. What do you get after you do that?
 

cgr4

New member
Aug 31, 2013
7
For the second question. It would look something like this?

125x3/2y3/2[HR][/HR]x2y

then I would subtract x2 and y

which would equal 125x1/2y ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
No, you want to apply the property of exponents:

\(\displaystyle \frac{a^b}{a^c}=a^{b-c}\)

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

cgr4

New member
Aug 31, 2013
7
Still a little confused about the 2nd problem.

So

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

correct?
 

cgr4

New member
Aug 31, 2013
7
resulting in 125x-1y

125y[HR]x[/HR]

125y over x
 

earboth

Active member
Jan 30, 2012
74
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:

\(\displaystyle 125 \cdot x^{-\frac12} \cdot y^{\frac12}\)

Re-write this term without the fractional exponents.