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A clear picture of the questions has been attatched. Please could you answer
)
11a and 11b and explain them to me please, thank you.
Hi Hassan,
a)let log4 3 = k, then
4^k = 3
(2^2)^k = 3
2^(2k) = 3
taking square roots,
2^k = sqrt3
and so,
log2 sqrt3 = k
therefore,
log4 3 = log2 sqrt3
b)2log2 y = log4 3 + log2 x
log2 y^2 = log2 sqrt3 + log2 x
log2 y^2 = log2 (sqrt3)x
y^2 = (sqrt3)x
3^y = 9^x
3^y = (3^2)^x
3^y = 3^(2x)
y = 2x
and so,
y^2 = 4x^2
comparing with the previous equation,
4x^2 = (sqrt3)x
4x^2 - (sqrt3)x = 0
x(4x - sqrt3) = 0
x = 0
y = 0
OR
4x - sqrt3 = 0
4x = sqrt3
x = (sqrt3)/4
y = (sqrt3)/2
Regards
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Ahmed Salami
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I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I can as well help a good deal in Physics with most emphasis directed towards mechanics.
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