Question

Solve the following systems of equations.
$lo{g}_{5}x+lo{g}_{5}7.log{}_{7}y=1+lo{g}_{5}2,3+2lo{g}_{2}y=lo{g}_{2}5(2+3lo{g}_{5}x).$

Answer 1

Solution: Given,

=> log₅ˣ + log₅⁷ .log₇ʸ = 1 + log₅²

Now, log₅ˣ + (logₑ⁷/logₑ⁵) x (logₑʸ/logₑ⁷) = 1 + log₅² [• logₐᵇ = logₑᵇ/logₑᵃ]

=> log₅ˣ + ( logₑʸ/ logₑ⁵) = log₅⁵ + log₅² [• logₐᵃ = 1]

=> log₅ˣ + log₅ʸ = log₅¹⁰ [• logₓᵃ + logₓᵇ = logₓᵃᵇ ]

=> log₅ˣʸ = log₅¹⁰ [• logₓᵃ = logₓᶜ , a= c]

=> xy = 10 equation → (1)

And, 3 + 2log₂ʸ = log₂⁵ (2 + 3log₅ˣ)

=> 3 + 2Log₂ʸ = 2log₂⁵ + 3log₂⁵ x log₅ˣ

=> 3 + 2Log₂ʸ = 2log₂⁵ + 3 x ( logₑ⁵/logₑ² ) x (logₑˣ/logₑ⁵) [•logₓᵃ = logₑᵃ/logₑˣ]

=> 3 + 2log₂ʸ = 2log₂⁵ + 3 x ( logₑˣ/logₑ²)

=> 3 + 2Log₂ʸ = 2 log₂⁵ + 3log₂ˣ [• logₑᵃ/logₑˣ = logₓᵃ]

=> 3 - 3Log₂ˣ = 2log₂⁵ - 2log₂ʸ

=> 3Log₂² - 3log₂ˣ = 2 ( log₂⁵ - log₂ʸ) [ • logₐᵃ = 1 ]

=> 3(log₂² - log₂ˣ ) = 2( log₂⁵ - log₂ʸ )

=> 3( log₂2/x ) = 2( log₂5/y ) [• logₓᵃ- logₓᵇ = logₓa/b ]

=> log₂(2/x)³ = log₂(5/y)² [• nlogₓᵃ = logₓaⁿ ]

=> (2/x)³ = (5/y)² [• logₓᵃ = logₓᶜ, a=c ]

=> 8/x³ = 25/y²

=> 8/x³ = 25/(10/x)² [• from equation (1) ,xy = 10]

=> 8/x³ = 25/(100/x²)

=> 8/x³ = 25x²/100

=> 8/x³ = x²/4

=> 32 = x⁵

=> 2⁵ = x⁵ [• aⁿ = bⁿ , a = b ]

=> x = 2 equation →(2)

• From equation → (1)

=> xy = 10

=> y = 10/x

=> y = 10/2 [ •from equation →(2) , x= 2]

=> y = 5

• Therefore,the solution of systems of equations are X = 2 , Y = 5