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secondary 3 | A Maths
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Please help I don't even know how to move on from there
Date Posted: 3 years ago
Views: 491
(logᵧx)² / logₓy + 125 = 0
(logᵧx)² × 1/logₓy = -125
(logᵧx)² × logᵧx = -125
(Recall that loga b = 1 / logb a)
(logᵧx)³ = -125
logᵧx = ³√-125
logᵧx = -5
x = y⁻⁵
x⁻¹/⁵ = y
(logᵧx)² × 1/logₓy = -125
(logᵧx)² × logᵧx = -125
(Recall that loga b = 1 / logb a)
(logᵧx)³ = -125
logᵧx = ³√-125
logᵧx = -5
x = y⁻⁵
x⁻¹/⁵ = y
Alternatively,
(logᵧx)² / logₓy + 125 = 0
(1/logₓy)² × 1/logₓy = -125
(1/logₓy)³ = -125
1/logₓy = ³√-125
1/logₓy = -5
logₓy = -1/5
y = x⁻¹/⁵
(logᵧx)² / logₓy + 125 = 0
(1/logₓy)² × 1/logₓy = -125
(1/logₓy)³ = -125
1/logₓy = ³√-125
1/logₓy = -5
logₓy = -1/5
y = x⁻¹/⁵
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Hi! Hope this helps! :)
Date Posted:
3 years ago
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