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If x/(x^(1. 5)\ )=8x^(-1) and x >0, then...

If `x/(x^(1. 5)\ )=8x^(-1)` and `x >0,` then `x=` `(sqrt(2))/4` (b) `2sqrt(2)` (c) 4 (d) 64

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`x/x^1.5=8x^(−1)`
`=>x/x^1.5=8/x`
`=>x^(1+1−1.5)=8`
`=>x^(2−1.5)=8`
`=>x^(0.5)=8`
`=>x^(1/2)=8`
Squaring both the sides:
`=>x^(1/2xx2)=8^2`
`=>x=64`
`therefore x=64`.
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