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secondary 4 | A Maths
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Just need part 2 pls
Date Posted: 5 years ago
Views: 262
Solve for u to get u=-1 or u=3
When u=-1, 2^x = -1 (no solution)... not possible to have lg(-1).
When u=3, 2^x =3,
x lg 2 = lg 3
x = (lg 3)/(lg 2)
When u=-1, 2^x = -1 (no solution)... not possible to have lg(-1).
When u=3, 2^x =3,
x lg 2 = lg 3
x = (lg 3)/(lg 2)
See 2 Answers
done
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clear
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For part (i), in each step I have used at least one specific law of logarithm up to the conversion into u.
Date Posted:
5 years ago
done
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clear
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in part i), indices laws are applied.
in part ii), factorisation of quadratic equation is done to find u, and the negative answer is rejected as 2^x cannot be negative. lg is then applied on both sides to find x. ln can also be used to get the same correct answer.
in part iii), the discriminant of the equation which works out to 4-4k is negative if k<-1, hence there will be no solution in that case.
in part ii), factorisation of quadratic equation is done to find u, and the negative answer is rejected as 2^x cannot be negative. lg is then applied on both sides to find x. ln can also be used to get the same correct answer.
in part iii), the discriminant of the equation which works out to 4-4k is negative if k<-1, hence there will be no solution in that case.
Date Posted:
5 years ago
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