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secondary 3 | A Maths
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can someone explain with steps pls ty !!
Date Posted: 5 years ago
Views: 408
given u = log₃ x
so 1/u = logₓ 3
[ bcos u = log₃ x = logx/log3 ;
& logₓ 3 = log3/logx
= 1 ÷ (logx/log3) = 1/u
(a)
log₃ x = 9 logₓ 3
=> u = 9 (1/u)
=> u = 9/u
=> u² = 9
=> u = 3 or -3
=> log₃ x = 3 or log₃ x = -3
=> x = 3³ or x = 3⁻³
=> x = 27 or x = 1/27
(b)
log₃ x + 2 = 3 logₓ 3
=> u + 2 = 3 (1/u)
=> u + 2 = 3/u
=> u² + 2u = 3
=> u² + 2u - 3 = 0
=> (u + 3)(u - 1) = 0
=> u = -3 or u = 1
=> log₃ x = -3 or log₃ x = 1
=> x = 3⁻³ or x = 3¹
=> x = 1/27 or x = 3
so 1/u = logₓ 3
[ bcos u = log₃ x = logx/log3 ;
& logₓ 3 = log3/logx
= 1 ÷ (logx/log3) = 1/u
(a)
log₃ x = 9 logₓ 3
=> u = 9 (1/u)
=> u = 9/u
=> u² = 9
=> u = 3 or -3
=> log₃ x = 3 or log₃ x = -3
=> x = 3³ or x = 3⁻³
=> x = 27 or x = 1/27
(b)
log₃ x + 2 = 3 logₓ 3
=> u + 2 = 3 (1/u)
=> u + 2 = 3/u
=> u² + 2u = 3
=> u² + 2u - 3 = 0
=> (u + 3)(u - 1) = 0
=> u = -3 or u = 1
=> log₃ x = -3 or log₃ x = 1
=> x = 3⁻³ or x = 3¹
=> x = 1/27 or x = 3
See 2 Answers
done
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Cheers
Date Posted:
5 years ago
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