Fractional exponents

[Richay]

These are pretty tough.

Here's three different ones

1.) 16^-1/2

2.) (-27)^-2/2

3.)

How do i solve them. I have no clue.

 

[Hockeyman]

16 ^ -1/2

would equal (1)/Sqrt 16 = 1/4

The denominator in your fraction exponent became your index of your radical, and since it was a negative exponent you must move it down to the denominator to make it positive.

Can you try the rest?

 

[Richay]

I understand how you did that one.

So, if i have to find the sqrt then of 27 it's 5.196......

Do i round it or am i doing this wrong?

 

[galactus]

Don't use decimals if you can get away with it.

\(\displaystyle \L\\\sqrt{27}=\sqrt{3}\sqrt{9}=3\sqrt{3}\)

See?.

 

[soroban]

Hello, Richay!

Exactly where is your difficulty?
$\displaystyle \;\;$You don't understand negative exponents?
$\displaystyle \;\;$You don't understand rational (fractional) exponents?
$\displaystyle \;\;$You can't do any algebra at all??

You're getting answers with decimals.
If you're going to use your calculator, what's the problem?

$\displaystyle 1)\;16^{-\frac{1}{2}}$

A negative exponent "moves" the expression.
$\displaystyle \;\;$If it is in the numerator, it moves to the denominator.
$\displaystyle \;\;$If it is in the denominator, it moves to the numerator.
And we drop the "minus".

So: $\displaystyle \;16^{-\frac{1}{2}} \;= \;\frac{1}{16^{\frac{1}{2}}}$

You're expected to know that a $\displaystyle \frac{1}{2}$ power means "square root".

So we have: $\displaystyle \;\frac{1}{\sqrt{16}}\:=\:\frac{1}{4}$

$\displaystyle 2)\;\frac{(-27)^{-2}}{2}$

We have: \(\displaystyle \L\;\frac{(-27)^{-2}}{2} \;= \;\frac{1}{2\,\cdot\,27^2}\;=\;\frac{1}{2\,\cdot\,729}\;=\;\frac{1}{1458}\)

$\displaystyle 3)\;\frac{1}{81^{-\frac{3}{4}}}$

This one gets "moved up": $\displaystyle \;\frac{1}{81^{-\frac{3}{4}}} \;= \;81^{\frac{3}{4}}$

You're expected to know that a $\displaystyle \frac{3}{4}$ power means: 3rd power, 4th root.

So we have: \(\displaystyle \:\sqrt[4]{81^3}\:=\:\left(\sqrt[4]{81}\right)^3\:=\3)^3\:=\:27\)

 

[Richay]

Don't use decimals if you can get away with it.

\(\displaystyle \L\\\sqrt{27}=\sqrt{3}\sqrt{9}=3\sqrt{3}\)

See?.

Yes. That helps me understand square roots better.

Hello, Richay!

Exactly where is your difficulty?
$\displaystyle \;\;$You don't understand negative exponents?
$\displaystyle \;\;$You don't understand rational (fractional) exponents?
$\displaystyle \;\;$You can't do any algebra at all??

You're getting answers with decimals.
If you're going to use your calculator, what's the problem?

$\displaystyle 1)\;16^{-\frac{1}{2}}$

A negative exponent "moves" the expression.
$\displaystyle \;\;$If it is in the numerator, it moves to the denominator.
$\displaystyle \;\;$If it is in the denominator, it moves to the numerator.
And we drop the "minus".

So: $\displaystyle \;16^{-\frac{1}{2}} \;= \;\frac{1}{16^{\frac{1}{2}}}$

You're expected to know that a $\displaystyle \frac{1}{2}$ power means "square root".

So we have: $\displaystyle \;\frac{1}{\sqrt{16}}\:=\:\frac{1}{4}$


[quote:3424z9uw]$\displaystyle 2)\;\frac{(-27)^{-2}}{2}$

We have: \(\displaystyle \L\;\frac{(-27)^{-2}}{2} \;= \;\frac{1}{2\,\cdot\,27^2}\;=\;\frac{1}{2\,\cdot\,729}\;=\;\frac{1}{1458}\)

$\displaystyle 3)\;\frac{1}{81^{-\frac{3}{4}}}$

This one gets "moved up": $\displaystyle \;\frac{1}{81^{-\frac{3}{4}}} \;= \;81^{\frac{3}{4}}$

You're expected to know that a $\displaystyle \frac{3}{4}$ power means: 3rd power, 4th root.

So we have: \(\displaystyle \:\sqrt[4]{81^3}\:=\:\left(\sqrt[4]{81}\right)^3\:=\3)^3\:=\:27\)[/quote:3424z9uw]

I see how you did those. The second one throw me off though.
If we change (-27)^-2/2 to: (-27)^-2/3 Then how would it be solved?

 

[Denis]

"If we change (-27)^-2/2 to: (-27)^-2/3 Then how would it be solved?"
Same way:
-27^(-2) / 3 = 1 / 3(-27^2) = 1 / 3(729) = 2187

What do you not understand? Remember: a^(-p) = 1 / a^p

 

[Richay]

I understand it almost 100% now, thanks for the help!