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Solve for x : 5^(x)root(x)(8^(x-1))=500...

Solve for x : `5^(x)root(x)(8^(x-1))=500`

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We have `5^(x)root(x)(8^(x-1))=5^(3).2^(2)`
`implies5^(x).7^(((x-1)/3))=5^(3).2^(2)`
`implies5^(x).2^((3x-3)/x)=5^(3).2^(2)`
`implies5^(x_3).2(((x-3)/x))=1`
`implies(5.2^(1//x)(x-3))=1`
is equivalent to the equation
`10^((x-3)log5.2^(1//x))=1`
`implies(x-3)log(5.2^(1//x))=0`
Thus, original equation is equivalent to the collection of equations
`x-3=0,log(5.2^(1//x))=0`
`:.x_(1)=3,5.2^(1//x)=1implies2^(1//x)=(1/5)`
`:.x_(2)=-log_(5)2`
Hence roots of the original equation are `x_(1)=3` and `x_(2)=-log_(5)2`.
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ARIHANT MATHS-THEORY OF EQUATIONS-Exercise (Questions Asked In Previous 13 Years Exam)
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