How do you express in terms of logarithms $log_b((p^2q^5)/(m^4b^9))$ ?

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Answer 1 · 2015-07-06T00:48:25

$log_b((p^2q^5)/(m^4b^9)) = 2log_bp + 5log_b(q) – 4log_bm - 9$

(1) $log_b(x/y) = log_bx -log_by$

So

$log_b((p^2q^5)/(m^4b^9)) = log_b(p^2q^5) – log_b(m^4b^9)$

(2) $log_b(xy) = log_bx – log_by$

So

$log_b((p^2q^5)/(m^4b^9)) = log_b(p^2) + log_b(q^5) – (log_b(m^4) + log_b(b^9))$

$log_b((p^2q^5)/(m^4b^9)) = log_b(p^2) + log_b(q^5) – log_b(m^4) - log_b(b^9)$

(3) $log_b(x^d) = dlog_bx$

So

$log_b((p^2q^5)/(m^4b^9)) = 2log_bp + 5log_b(q) – 4log_bm - 9log_b b$

$log_b((p^2q^5)/(m^4b^9)) = 2log_bp + 5log_b(q) – 4log_bm - 9$

How do you express in terms of logarithms log_b((p^2q^5)/(m^4b^9))log_b((p^2q^5)/(m^4b^9))?