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प्रश्न
Find the value of x in the following:
`(root3 4)^(2x+1/2)=1/32`
उत्तर
Given `(root3 4)^(2x+1/2)=1/32`
`(2^2)^((1/3)((4x+1)/2))=(1/2)^5`
`rArr2^((4x+1)/3)=2^-5`
On comparing we get,
`(4x+1)/3=-5`
⇒ 4x + 1 = -5 x 3
⇒ 4x + 1 = -15
⇒ 4x = -15 - 1
⇒ 4x = -16
`rArrx=-16/4`
⇒ x = -4
Hence, the value of x = -4.
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