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secondary 3 | A Maths
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pls help
Date Posted: 3 years ago
Views: 197
5 log2 y - logy 2 = 4
5 log2 y - 1 / log2 y = 4
(Property used is : loga b = 1 / logb a)
Now use the substitution x = log2 y
5x - 1/x = 4
5x² - 1 = 4x
(multiply whole equation by x)
5x² - 4x - 1 = 0
(5x + 1)(x - 1) = 0
5x + 1 = 0 or x - 1 = 0
5x = -1 or x = 1
x = -1/5
So,
log2 y = -1/5 or log2 y = 1
y = 2^(-1/5) or y = 2¹
y = 1/2^(1/5) or y = 2
y = 1/⁵√2
So y = 1/⁵√2 or y = 2
5 log2 y - 1 / log2 y = 4
(Property used is : loga b = 1 / logb a)
Now use the substitution x = log2 y
5x - 1/x = 4
5x² - 1 = 4x
(multiply whole equation by x)
5x² - 4x - 1 = 0
(5x + 1)(x - 1) = 0
5x + 1 = 0 or x - 1 = 0
5x = -1 or x = 1
x = -1/5
So,
log2 y = -1/5 or log2 y = 1
y = 2^(-1/5) or y = 2¹
y = 1/2^(1/5) or y = 2
y = 1/⁵√2
So y = 1/⁵√2 or y = 2
See 1 Answer
done
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clear
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Here the “negative” value arising from the quadratic equation should not be rejected.
Date Posted:
3 years ago
Can just leave in exact form actually.
i.e y = 1/⁵√2 or y = 2
i.e y = 1/⁵√2 or y = 2
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