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erastovalidia [21]
3 years ago
8

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

Mathematics
1 answer:
anyanavicka [17]3 years ago
3 0

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}

♨Rage♨

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Step-by-step explanation:

Question 1:

We need to rewrite the expression using exponents

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