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secondary 3 | A Maths
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Date Posted: 3 years ago
Views: 181
You missed out a negative sign for log3 3^y in your third line for ⑵
Can I ask, where is this question from?
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Your working for the first equation is correct (y = 5x - 1)
The second equation should be read as such (if not, it is not solvable at your level. The setter should have included brackets for clarity)
log₃(1 - y) - 2 log₃x = 1
Then,
log₃(1 - y) - log₃x² = log₃3
log₃(1 - y) = log₃3 + log₃x²
log₃(1 - y) = log₃(3x²)
1 - y = 3x²
Sub y = 5x - 1,
1 - (5x - 1) = 3x²
2 - 5x = 3x²
3x² + 5x - 2 = 0
Factorise,
(3x - 1)(x + 2) = 0
3x - 1 = 0 or x + 2 = 0
3x = 1 or x = -2 (rejected as that would make 2 log₃x undefined)
x = ⅓
So, y = 5(⅓) - 1 = ⅔
The second equation should be read as such (if not, it is not solvable at your level. The setter should have included brackets for clarity)
log₃(1 - y) - 2 log₃x = 1
Then,
log₃(1 - y) - log₃x² = log₃3
log₃(1 - y) = log₃3 + log₃x²
log₃(1 - y) = log₃(3x²)
1 - y = 3x²
Sub y = 5x - 1,
1 - (5x - 1) = 3x²
2 - 5x = 3x²
3x² + 5x - 2 = 0
Factorise,
(3x - 1)(x + 2) = 0
3x - 1 = 0 or x + 2 = 0
3x = 1 or x = -2 (rejected as that would make 2 log₃x undefined)
x = ⅓
So, y = 5(⅓) - 1 = ⅔
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