Question

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Given log x = 3 and log y = 2, prove that x^2 -7xy =3000y​​

Answer 1

`if \: log(x ) = 3 \\ log_(10)(x) = 3 \\ {10}^(3) = x \\ therfore \to \: \boxed{x = {10}^(3) } \\ \\ if \: log(y ) = 2 \\ log_(10)(y) = 2 \\ {10}^(2) = y \\ therfore \to \: \boxed{y= {10}^(2) }`

Explanation:

`x^2 -7xy =3000y \: then \to \\ x^2 =3000y + 7xy \\ {(10 ^(3) )}^(2) = 3000(10)^(2) + 7( {10}^(3) * {10}^(2)) \\ {10}^(6) = 3000(10)^(2) + 7( {10}^(3) * {10}^(2)) \\ {10}^(6) = 3000(10)^(2) + 7( {10}^(5) ) \\ {10}^(6) = 300000 + 700000 \\ \boxed{{10}^(6) = 1000000} \\ hence \: the \: equation \to \: x^2 -7xy =3000y \\ \boxed{ is \: corret}`

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