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anonymous
 one year ago
Expand the logarithm
Quesition about placement 
My question is do I need to add parenthesis to logarithm answers?
anonymous
 one year ago
Expand the logarithm Quesition about placement  My question is do I need to add parenthesis to logarithm answers?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My book has these all over the place so I'm not sure.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you should add parenthesis idk i might be wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the "2" is multiplying the logarithmic function, thus if the logarithm expands to the "sum" version, the "2" has to multiply the expanded logarithmic version then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so then yes add the parenthesis?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf log_4(3xyz)^2\implies 2\left( log_43+log_4x+log_4y+log_4z \right) \\ \quad \\ 2log_43+2log_4x+2log_4y+2log_4z\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay you're confusing me you just wrote both down

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm nope, the last line is the parenthesized version, expanded

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so I don't add parenthesis.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well.. distributing the "2" will be, expanding, thus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm confused now and idk what I'm doing

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the first step is to pull down the exponent 2 using this rule \[\Large \log_{b}(x^y) = y*\log_b(x)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1after you pulled down the 2, you will have \[\Large 2\log_{4}\left(3xyz\right)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then you'll use the rule log(x*y) = log(x) + log(y) to expand out log(3xyz)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1use the rule log(x*y) = log(x) + log(y) to get... \[\Large 2\log_{4}\left(3xyz\right)\] \[\Large 2[\log_{4}\left(3xyz\right)]\] \[\Large 2[\log_{4}(3)+\log_{4}(xyz)]\] \[\Large 2[\log_{4}(3)+\log_{4}(x)+\log_{4}(yz)]\] \[\Large 2[\log_{4}(3)+\log_{4}(x)+\log_{4}(y)+\log_{4}(z)]\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I used the rule in the smallest pieces possible. So it's a bit more dragged out than it has to be. You can just go from log(3xyz) to log(3)+log(x)+log(y)+log(z) in one step really

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1anyways, the point is that 2 on the outside is being multiplied by each term inside. You can leave the 2 out there like it is, but make sure you have surrounding parenthesis around those log terms don't write \[\Large 2\log_{4}(3)+\log_{4}(x)+\log_{4}(y)+\log_{4}(z)\] instead write \[\Large 2[\log_{4}(3)+\log_{4}(x)+\log_{4}(y)+\log_{4}(z)]\] or write \[\Large 2(\log_{4}(3)+\log_{4}(x)+\log_{4}(y)+\log_{4}(z))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah I have options to choose from and the parenthesis around the 3,x,y,and z isn't an option it's just 2log4 3+log4 x+log4 y +log4 z or 2(log4 3+log4 x+log4 y +log4 z) so I feel like I will just go w/ the parenthesis.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then go with 2(log4 3+log4 x+log4 y +log4 z)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my book has examples written like yours so i think that's why I was confused. Thank you for helping me

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1as for what jdoe0001 did, he simply used the distribution rule to multiply the '2' by each term inside eg: 2*(x+y) = 2*x + 2*y

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm glad I could help clear things up
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