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secondary 3 | A Maths
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help with dis pls, thank you in advance
Date Posted: 3 years ago
Views: 538
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lmk if u don’t understand or any issue with the answer :)
Date Posted:
3 years ago
ans is (ab/1-a) tho
The answer has to fully be in terms of a and b only.
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hope this helps
Date Posted:
3 years ago
ans is (ab/1-a) tho
Your log6 3 can be further simplified.
Given : log₆2 = a, log₅3 = b
To find : log₅2
①log₅2
= log₅(6 ÷ 3)
= log₅6 - log₅3
= (log₅2)/(log₆2) - b
(Recall that logₐb = log꜀b / log꜀a = (1/logb c) / (1/logₐc) = logₐc / logb c)
= (log₅2)/a - b
②Now we have log₅2 on both sides. Multiply both sides by a to get rid of the denominator.
a log₅2 = log₅2 - ab
③Bring over ab to the left and a log₅2 to the right.
ab = log₅2 - a log₅2
ab = (1 - a)(log₅2)
log₅2 = ab/(1 - a)
To find : log₅2
①log₅2
= log₅(6 ÷ 3)
= log₅6 - log₅3
= (log₅2)/(log₆2) - b
(Recall that logₐb = log꜀b / log꜀a = (1/logb c) / (1/logₐc) = logₐc / logb c)
= (log₅2)/a - b
②Now we have log₅2 on both sides. Multiply both sides by a to get rid of the denominator.
a log₅2 = log₅2 - ab
③Bring over ab to the left and a log₅2 to the right.
ab = log₅2 - a log₅2
ab = (1 - a)(log₅2)
log₅2 = ab/(1 - a)
thank you!!
Alternative working :
log₆2 = a, log₅3 = b
log₅2
= log₆2 / log₆5
= a × 1/log₆5
= a log₅6
= a log₅(3 × 2)
= a (log₅3 + log₅2)
= a(b + log₅2)
= ab + a log₅2
ab = log₅2 - a log₅2
ab = (1 - a)(log₅2)
log₅2 = ab/(1 - a)
log₆2 = a, log₅3 = b
log₅2
= log₆2 / log₆5
= a × 1/log₆5
= a log₅6
= a log₅(3 × 2)
= a (log₅3 + log₅2)
= a(b + log₅2)
= ab + a log₅2
ab = log₅2 - a log₅2
ab = (1 - a)(log₅2)
log₅2 = ab/(1 - a)
Second alternative :
log₆2 = a
log₅2 / log₅6 = a
log₅2 / log₅(2 × 3) = a
log₅2 / (log₅2 + log₅3) = a
log₅2 / (log₅2 + b) = a
log₅2 = a(log₅2 + b)
log₅2 = a log₅2 + ab
log₅2 - a log₅2 = ab
(1 - a) log₅2 = ab
log₅2 = ab / (1 - a)
Third alternative : indices way
log₆2 = a
2 = 6ᵃ
2 = (2×3)ᵃ = (2ᵃ)(3ᵃ)
2 / 2ᵃ = 3ᵃ
2¹⁻ᵃ = 3ᵃ ①
log₅3 = b
3 = 5ᵇ
3ᵃ = (5ᵇ)ᵃ
3ᵃ = 5ᵃᵇ ②
Comparing ① and ②,
2¹⁻ᵃ = 5ᵃᵇ
Take log₅ on both sides,
log₅2¹⁻ᵃ = log₅5ᵃᵇ
(1 - a) log₅2 = ab log₅5
(1 - a) log₅2 = ab
log₅2 = ab / (1 - a)
log₆2 = a
log₅2 / log₅6 = a
log₅2 / log₅(2 × 3) = a
log₅2 / (log₅2 + log₅3) = a
log₅2 / (log₅2 + b) = a
log₅2 = a(log₅2 + b)
log₅2 = a log₅2 + ab
log₅2 - a log₅2 = ab
(1 - a) log₅2 = ab
log₅2 = ab / (1 - a)
Third alternative : indices way
log₆2 = a
2 = 6ᵃ
2 = (2×3)ᵃ = (2ᵃ)(3ᵃ)
2 / 2ᵃ = 3ᵃ
2¹⁻ᵃ = 3ᵃ ①
log₅3 = b
3 = 5ᵇ
3ᵃ = (5ᵇ)ᵃ
3ᵃ = 5ᵃᵇ ②
Comparing ① and ②,
2¹⁻ᵃ = 5ᵃᵇ
Take log₅ on both sides,
log₅2¹⁻ᵃ = log₅5ᵃᵇ
(1 - a) log₅2 = ab log₅5
(1 - a) log₅2 = ab
log₅2 = ab / (1 - a)
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