C2 Logarthms Questions

Log3 (2 - 3x) = log9 (6x² - 19x + 2)

log[9]x = n
=> x = 9^n = 3^(2n)
=> log[3]x = 2n = 2.log[9]x
=> log[3](2 - 3x) = 2.log[9](2-3x) = log[9](2-3x)²
=> (2-3x)² = (6x² - 19x + 2)

4 - 12x + 9x² = 6x² - 19x + 2
3x² + 7x + 2 = 0
(3x + 1)(x + 2) = 0
x = -1/3, x= -2
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If xy=64 and logxy + logyx = 5/2, find x and y.

log[y]x = 1/log[x]y

log[x]y + 1/log[x]y = 5/2
u + 1/u = 5/2, letting u = log[x]y
2u² - 5u + 2 = 0
(2u-1)(u-2) = 0
u = ½, u = 2

log[x]y = ½ => x = y^½ or y = x²
log[x]y = 2 => x = y²

let x = y²

xy = 64
y^3 = 64
y = 4, x = 16
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