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AnimalAin
 2 years ago
Best ResponseYou've already chosen the best response.1The rules of logarithms are important here:\[\log(AB)=logA+logB\]\[\log(\frac{A}{B})=logAlogB\]

Cutiepo0
 2 years ago
Best ResponseYou've already chosen the best response.0I got as far as log (((a^2)+ab)/(c/d))

Cutiepo0
 2 years ago
Best ResponseYou've already chosen the best response.0but in the answer, the bottom of the equation is cd not c/d and I don't get why :(

AnimalAin
 2 years ago
Best ResponseYou've already chosen the best response.1Working from that,\[\log a + \log (a+b)  \log c  \log d=\log(\frac{a(a+b)}{cd})\]

AnimalAin
 2 years ago
Best ResponseYou've already chosen the best response.1OK, this time stepbystep. From the product rule\[\log a + \log (a+b)  \log c  \log d=\log[a(a+b)]\log c\log d\]

AnimalAin
 2 years ago
Best ResponseYou've already chosen the best response.1Now, from the quotient rule\[\log[a(a+b)]−logc−logd=\log[\frac{a(a+b)}{c}]logd\]

AnimalAin
 2 years ago
Best ResponseYou've already chosen the best response.1Apply the quotient rule again, to get\[\log[\frac{a(a+b)}{c}]−logd=\log[\frac{a(a+b)}{cd}]\]

Cutiepo0
 2 years ago
Best ResponseYou've already chosen the best response.0and then with log d you do the same step as the last time!! Oh! How did you know to do the c first and then d though, I just did the first two parts with the multiplication law, and thelast two with the division law, and then both together with the division law

AnimalAin
 2 years ago
Best ResponseYou've already chosen the best response.1You eat an elephant one bite at a time. Do math the same way.

Cutiepo0
 2 years ago
Best ResponseYou've already chosen the best response.0haha okay thanks so much :)
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