Jasper Eng's answer to sunlight's Secondary 3 A Maths Singapore question.
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Log4 a = x
Log2 a / log2 4 = x
1/2 * log2 (a) = x (4 = 2^ 2, Log2 2 = 1)
Log2 a = 2x -------(1)
Log64 b = y
Log2 b / log2 64 = y (64 = 2^6)
1/6 * log2 (b)= y
Log2 b = 6y --------(2)
a/b = 2^z
z = log2 (a/b)
= log2 a - log2 b
= 2x - 6y //
Log2 a / log2 4 = x
1/2 * log2 (a) = x (4 = 2^ 2, Log2 2 = 1)
Log2 a = 2x -------(1)
Log64 b = y
Log2 b / log2 64 = y (64 = 2^6)
1/6 * log2 (b)= y
Log2 b = 6y --------(2)
a/b = 2^z
z = log2 (a/b)
= log2 a - log2 b
= 2x - 6y //
Date Posted:
3 years ago
Don't really understand clearly
Answer is correct. Revise your logarithm laws.
Ya I got it already just change the log to exponential form then solve
Thanks J can help with another qns
Use change of base law,
to change all base to same value, which in this case is 2.
Value 4 and 64 are powers of 2, hence can be evaluated in logarithm forms.
Change of Base:
LogA (b) = LogC (A) / LogC (b),
Where C can be chosen any appropriate value
to change all base to same value, which in this case is 2.
Value 4 and 64 are powers of 2, hence can be evaluated in logarithm forms.
Change of Base:
LogA (b) = LogC (A) / LogC (b),
Where C can be chosen any appropriate value
Thanks
It's actually logₐ b = log꜀ b / log꜀ a
Or
logₐ b = logₐ c / logb c
Or
logₐ b = logₐ c / logb c
Oops, yeap mistake** follow his
Okay
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