Question

Solve the following systems of equations.
$3.{\frac{2}{3}}^{2x-y}+7.\frac{2}{3}-6=0,log(3x-y)+log(y+x)=4log2.$

Answer 1

Solution: Given,

=> 3•(2²ˣ⁻ʸ)/3 + 7•2/3 - 6 = 0

=> 3•(2²ˣ⁻ʸ)/3 = 6 - 14/3

=> 3•( 2²ˣ⁻ʸ )/3 = 4/3

• Taking log both side having base 10 , we have

=> log (3•( 2²ˣ⁻ʸ )/3) = log4/3

=> log2²ˣ⁻ʸ = log4/3

=> ( 2x -y )log2 = log4/3 [• logaⁿ = nloga ]

=> 2( 2x - y ) = 4/3 [• loga = logb , a=b ]

=> 2x - y = 2/3

=> 3( 2x - y ) = 2

=> 6x - 3y= 2

=> y = ( 6x - 2 )/3 - equation→(1)

And, log( 3x - y ) + log ( y + x) = 4log2

=> log(3x - y)( y + x ) = log2⁴ [• loga + logb = logab and nloga = logaⁿ ]

=> ( 3x - y )( x + y ) = 2⁴ [• log a = logb , a = b ]

=> 3x² - xy + 3xy - y² = 16

=>3x² + 2xy - y² = 16

=> 3x² + 2x•( 6x - 2 )/3 - ( 6x - 2 )²/3² = 16 [• y value from equation (1) ]

=> 3x² + ( 12x² - 4x )/3 - (36x² + 4 - 24x )/9 = 16

=> [ 27x² + 3(12x² - 4x ) - ( 36x² + 4 - 24x )] /9 = 16

=> 27x² + 36x²- 12x - 36x²- 4 + 24x =144

=> 27x² + 12x = 148

•This is quadratic equation ,

so, determinant of the equation is given by

=> D = b²- 4ac

=> D = 12² - 4x27 (-148)

=> D = 144 + 15984

=> D = 16128

And, √d = 48√7

• so, solution of quadratic equation is given as

=> x = ( b ± √D )/2a

=> x = ( 12 ± 48√7 ) / 2 x 27

=> x = 12 ( 1 ± 4√7 )/ 2 x 27

=> x = 2 ( 1 ± 4√7 ) / 9

=> x = ( 2 ± 8√7 ) / 9

• But argument of log cannot be negative

so, X = ( 2 + 8√7 ) / 9 -(2)

From equation (1) ,we have

=> y = ( 6x - 2 )/3

=> y=[ 6 ( 2 + 8√7 )/9 -2 ]/3 [• from equation (2), x = ( 2 + 8√7 )/9 ]

=> y = [ 2 (2 + 8√7 ) - 6 ]/9

=> y = ( 4 + 16√7 - 6 )/9

=> y = ( 16√7 - 2 )/9

•Therefore solution of systems of equations are

X = ( 2 + 8√7 )/9 , Y= ( 16√7 - 2 )/9