Solved on Sep 23, 2023

Simplify the expression $9^{-\frac{1}{2}} \times \frac{1}{27}$ and show it is equal to $3^{-4}$ .

STEP 1

Assumptions1. We are using the properties of exponents, which state that $a^{-n} = \frac{1}{a^n}$ , $(a^n)^m = a^{n \times m}$ and $\frac{a^n}{a^m} = a^{n-m}$ . . We are using the fact that 9 =3^ and $27 =3^3$ .

STEP 2

First, we rewrite $9^{-\frac{1}{2}}$ and $\frac{1}{27}$ in terms of base.
$9^{-\frac{1}{2}} = (^2)^{-\frac{1}{2}}$ $\frac{1}{27} =^{-}$

STEP 3

Now, we apply the property of exponents $(a^n)^m = a^{n \times m}$ to simplify $(3^2)^{-\frac{1}{2}}$ .
$(3^2)^{-\frac{1}{2}} =3^{2 \times -\frac{1}{2}}$

STEP 4

Calculate the exponent.
$3^{2 \times -\frac{1}{2}} =3^{-1}$

STEP 5

Now, we multiply $3^{-1}$ and $3^{-3}$ .
$3^{-1} \times3^{-3} =3^{-1-3}$

STEP 6

Calculate the exponent.
$3^{-1-3} =3^{-4}$ So, $9^{-\frac{1}{2}} \times \frac{1}{27}=3^{-4}$ .

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Simplify the expression $9^{-\frac{1}{2}} \times \frac{1}{27}$ and show it is equal to $3^{-4}$ .

STEP 1

Assumptions1. We are using the properties of exponents, which state that $a^{-n} = \frac{1}{a^n}$ , $(a^n)^m = a^{n \times m}$ and $\frac{a^n}{a^m} = a^{n-m}$ . . We are using the fact that 9 =3^ and $27 =3^3$ .

STEP 2

First, we rewrite $9^{-\frac{1}{2}}$ and $\frac{1}{27}$ in terms of base.
$9^{-\frac{1}{2}} = (^2)^{-\frac{1}{2}}$ $\frac{1}{27} =^{-}$

STEP 3

Now, we apply the property of exponents $(a^n)^m = a^{n \times m}$ to simplify $(3^2)^{-\frac{1}{2}}$ .
$(3^2)^{-\frac{1}{2}} =3^{2 \times -\frac{1}{2}}$

STEP 4

Calculate the exponent.
$3^{2 \times -\frac{1}{2}} =3^{-1}$

STEP 5

Now, we multiply $3^{-1}$ and $3^{-3}$ .
$3^{-1} \times3^{-3} =3^{-1-3}$

STEP 6

Calculate the exponent.
$3^{-1-3} =3^{-4}$ So, $9^{-\frac{1}{2}} \times \frac{1}{27}=3^{-4}$ .

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Simplify the expression 9−12×1279^{-\frac{1}{2}} \times \frac{1}{27}9−21​×271​ and show it is equal to 3−43^{-4}3−4.

STEP 1

Assumptions1. We are using the properties of exponents, which state that a−n=1ana^{-n} = \frac{1}{a^n}a−n=an1​, (an)m=an×m(a^n)^m = a^{n \times m}(an)m=an×m and anam=an−m\frac{a^n}{a^m} = a^{n-m}aman​=an−m. . We are using the fact that 9 =3^ and 27=3327 =3^327=33.

STEP 2

First, we rewrite 9−129^{-\frac{1}{2}}9−21​ and 127\frac{1}{27}271​ in terms of base.9−12=(2)−129^{-\frac{1}{2}} = (^2)^{-\frac{1}{2}}9−21​=(2)−21​127=−\frac{1}{27} =^{-}271​=−

STEP 3

Now, we apply the property of exponents (an)m=an×m(a^n)^m = a^{n \times m}(an)m=an×m to simplify (32)−12(3^2)^{-\frac{1}{2}}(32)−21​.(32)−12=32×−12(3^2)^{-\frac{1}{2}} =3^{2 \times -\frac{1}{2}}(32)−21​=32×−21​

STEP 4

Calculate the exponent.32×−12=3−13^{2 \times -\frac{1}{2}} =3^{-1}32×−21​=3−1

STEP 5

Now, we multiply 3−13^{-1}3−1 and 3−33^{-3}3−3.3−1×3−3=3−1−33^{-1} \times3^{-3} =3^{-1-3}3−1×3−3=3−1−3

STEP 6

Calculate the exponent.3−1−3=3−43^{-1-3} =3^{-4}3−1−3=3−4So, 9−12×127=3−49^{-\frac{1}{2}} \times \frac{1}{27}=3^{-4}9−21​×271​=3−4.

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Simplify the expression $9^{-\frac{1}{2}} \times \frac{1}{27}$ and show it is equal to $3^{-4}$ .

Assumptions1. We are using the properties of exponents, which state that $a^{-n} = \frac{1}{a^n}$ , $(a^n)^m = a^{n \times m}$ and $\frac{a^n}{a^m} = a^{n-m}$ . . We are using the fact that 9 =3^ and $27 =3^3$ .

First, we rewrite $9^{-\frac{1}{2}}$ and $\frac{1}{27}$ in terms of base.
$9^{-\frac{1}{2}} = (^2)^{-\frac{1}{2}}$ $\frac{1}{27} =^{-}$

Now, we apply the property of exponents $(a^n)^m = a^{n \times m}$ to simplify $(3^2)^{-\frac{1}{2}}$ .
$(3^2)^{-\frac{1}{2}} =3^{2 \times -\frac{1}{2}}$

Calculate the exponent.
$3^{2 \times -\frac{1}{2}} =3^{-1}$

Now, we multiply $3^{-1}$ and $3^{-3}$ .
$3^{-1} \times3^{-3} =3^{-1-3}$

Calculate the exponent.
$3^{-1-3} =3^{-4}$ So, $9^{-\frac{1}{2}} \times \frac{1}{27}=3^{-4}$ .