O lvl A Maths: Indices and Logarithms

From http://www.sgforums.com/forums/2297/topics/332604 courtesy of Ahm97sic

(1) Evaluate log_n 9 / log_n 4

(2) Given that zy = 2^{[log₂ (z + x)]}, show that z = x / (y-1)

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Answer:

(1)
log_n 9 / log_n 4
= (lg 9 / lg n) / (lg 4 / lg n)
= lg 9 / lg 4
= (2 * lg 3) / (2 * lg 2)
= lg 3 / lg 2
= 1.585

(2)
zy = 2^{[log₂ (z + x)]}
zy = z + x ===> 2^{[log₂ x]} = x, or n^{[log_n x]} = x
zy - z = x
z(y - 1) = x
z = x/(y - 1) (shown)

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