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

Question

2842 views around the world

You can reuse this answer
Creative Commons License

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#