Question

REALLY APPRECIATE YOUR HELPGiven log x = 3 and log y = 2, prove that x^2 -7xy =3000y​​

Answer 1

Answer:

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

Step-by-step explanation:

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

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