Question

решите систему уравнений с логарифмами(сложно).СПААААСИТЕ МЕНЯ

Answer 1

$left{begin{array}{l} log^3_4{ y^frac{1}{3} } - ( frac{1}{3} )^{-3x} = -9 , \\ log^2_4{y} + ( frac{1}{3} )^{-x} cdot log_4{ y^3 } = 27 - 9^{x+1} ; end{array}right$

$left{begin{array}{l} ( frac{1}{3} log_4{y} )^3 - 3^{3x} = -3^2 , \\ log^2_4{y} + 3 cdot 3^x cdot log_4{y} = 3^3 - 9 cdot 3^{2x} ; end{array}right$

$left{begin{array}{l} frac{1}{3^3} ( log_4{y} )^3 - ( 3^x )^3 = -3^2 , \\ log^2_4{y} + ( 3 cdot 3^x ) cdot log_4{y} + (3 cdot 3^x)^2 = 3^3 ; end{array}right$

$left{begin{array}{l} ( log_4{y} )^3 - ( 3 cdot 3^x )^3 = -3^5 , \ ( log_4{y} )^2 + ( 3 cdot 3^x ) cdot log_4{y} + (3 cdot 3^x)^2 = 3^3 ; end{array}right$

$left{begin{array}{l} u = log_4{y} , \ k = 3 cdot 3^x , \ u^3 - k^3 = -3^5 , \ u^2 + uk + k^2 = 3^3 ; end{array}right$

$left{begin{array}{l} u = log_4{y} , k = 3 cdot 3^x , \ ( u - k ) ( u^2 + uk + k^2 ) = -3^5 , \ u^2 + uk + k^2 = 3^3 ; end{array}right$

$left{begin{array}{l} u = log_4{y} , k = 3 cdot 3^x , \ ( u - k ) cdot 3^3 = -3^5 , \ u^2 + uk + k^2 = 3^3 ; end{array}right$

$left{begin{array}{l} u = log_4{y} , k = 3 cdot 3^x , \ u - k = -3^2 , \ u^2 + uk + k^2 = 3^3 ; end{array}right$

$left{begin{array}{l} u = log_4{y} , k = 3 cdot 3^x , \ u = k - 3^2 , \ left|begin{array}{l} ( k - 3^2 )^2 + k ( k - 3^2 ) + k^2 = 3^3 ; \ k^2 - 2 cdot 3^2 k + 3^4 + k^2 - 3^2 k + k^2 = 3^3 ; \ 3 k^2 - 3 cdot 3^2 k + 3^4 = 3^3 ; \ k^2 - 9 k + 18 = 0 ; \ D = 9^2 - 4 cdot 18 = 9 = 3^2 ; \ k_{1,2} = frac{ 9 pm 3 }{2} ; end{array}right end{array}right$

$left{begin{array}{l} k_{1,2} in { 3 , 6 } , \ u_{1,2} in { -6 , -3 } , \ 3 cdot 3^{ x_{1,2} } in { 3 , 6 } , \ log_4{ y_{1,2} } in { -6 , -3 } ; end{array}right$

$left{begin{array}{l} 3^x_{1,2} in { 1 , 2 } , \ y_{1,2} in { 4^{-6} , 4^{-3} } ; end{array}right$

$( x ; y ) in { ( 0 ; frac{1}{4096} ) , ( log_3{2} ; frac{1}{64} ) }$ ;


О т в е т : ${ ( 0 ; frac{1}{4096} ) , ( log_3{2} ; frac{1}{64} ) } .$