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SYSTEM:
log base root3 of (x-y) = 2

log base 4 of x - log base x of y = 7/6

User LinconFive
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1 Answer

4 votes

Answer:

Explanation:

SYSTEM:

log base root3 of (x-y) = 2 .... 1

log base 4 of x - log base x of y =7/6...2

From equation 1:

x-y = (√3)²

x - y = 3 ..... 3

From equation 2:

log base 4 of x - log base x of y =7/6

-log base x of 4 - log base x of y = 7/6

log base x of 4^-1 - log base x of y = 7/6

log base x of (1/4) - log base x of y = 7/6

According to law of logarithm

Log base x {1/4 ÷ y} = 7/6

x^7/6 = 1/4y .... 2

From 1, y = x-3

x^7/6 =1/4(x-3)

x^7/6 = 1/(4x-12)

x^7/6(4x-12) = 1

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User Dacology
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