Answer:
Step-by-step explanation:
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Assuming
is a non-negative real number...
Raise both sides to the power 2 :
![a^{3/4} = 8 \implies \left(a^{3/4}\right)^2 = 8^2 \implies a^{3/2} = 64](https://tex.z-dn.net/?f=a%5E%7B3%2F4%7D%20%3D%208%20%5Cimplies%20%5Cleft%28a%5E%7B3%2F4%7D%5Cright%29%5E2%20%3D%208%5E2%20%5Cimplies%20a%5E%7B3%2F2%7D%20%3D%2064)
which follows from the exponent product property,
. In particular, 3/4 × 2 = 6/4 = 3/2.
Raise both sides to the power 1/3 :
![\left(a^{3/2}\right)^{1/3} = 64^{1/3} \implies a^{1/2} = 4](https://tex.z-dn.net/?f=%5Cleft%28a%5E%7B3%2F2%7D%5Cright%29%5E%7B1%2F3%7D%20%3D%2064%5E%7B1%2F3%7D%20%5Cimplies%20a%5E%7B1%2F2%7D%20%3D%204)
where we use the same property as before, and 4³ = 64. This time, 3/2 × 1/3 = 3/6 = 1/2.
Take the reciprocal of both sides. This negates the exponent, so we end up with
![\dfrac1{a^{1/2}} = \dfrac14 \implies \boxed{a^{-1/2} = \dfrac14}](https://tex.z-dn.net/?f=%5Cdfrac1%7Ba%5E%7B1%2F2%7D%7D%20%3D%20%5Cdfrac14%20%5Cimplies%20%5Cboxed%7Ba%5E%7B-1%2F2%7D%20%3D%20%5Cdfrac14%7D)
Of course, you could also solve for
immediately by raising both sides of the original equation to the power 4/3. We have 4/3 × 3/4 = 12/12 = 1, so
![a^{3/4} = 8 \implies \left(a^{3/4}\right)^{4/3} = 8^{4/3} \implies a = 8^{4/3}](https://tex.z-dn.net/?f=a%5E%7B3%2F4%7D%20%3D%208%20%5Cimplies%20%5Cleft%28a%5E%7B3%2F4%7D%5Cright%29%5E%7B4%2F3%7D%20%3D%208%5E%7B4%2F3%7D%20%5Cimplies%20a%20%3D%208%5E%7B4%2F3%7D)
Now 2³ = 8, so
, and since 4² = 16, it follows that
![a = 16 \implies a^{1/2} = \sqrt{16} = 4 \implies a^{-1/2} = \dfrac14](https://tex.z-dn.net/?f=a%20%3D%2016%20%5Cimplies%20a%5E%7B1%2F2%7D%20%3D%20%5Csqrt%7B16%7D%20%3D%204%20%5Cimplies%20a%5E%7B-1%2F2%7D%20%3D%20%5Cdfrac14)
Answer:
1. y/4 = 15/20 or (y times 5)/20 = 15/20
Y = 3
Step-by-step explanation:
First, make sure the denominators are the same and then it will be up to multiplication and division.
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Step-by-step explanation:ees
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