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 2 years ago
Simplify the expression attached and write it as a single logarithm..
***still confused on this topic... pls help? :)
 2 years ago
Simplify the expression attached and write it as a single logarithm.. ***still confused on this topic... pls help? :)

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iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1answer choices A,B,C,D from top to bottom :)

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Move all the exponents up: \((x+4)^{3}\) \((x7)^{2}\) \((x2)^{5}\) \((x^{2})^{1}\) Notice how I deliberately left everything outside as ADDITION. This allows us just to bring everything inside without concern for numerator or denominator. We can figure all that out later.

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1im confused.. so how can i apply that with my expression?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Confused with what? \(3\log(x+4) = \log\left[\left(x+4\right)^{3}\right]\) Correct?

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1im not quite sure what we are working on now.. I'm like uber confused.. are we doing my problem? or an example? :/ sorry I'm a bit confused :(

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Your problem statement begins \(3\log(x+4)\), doesn't it? Or am I looking at some other picture?

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1yup iy does :) so i get log[(x+4)^−3] from that part?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Very good. Now the (x7) part?

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.12log(x7) = log(x7)^2 ??

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.15log(x2)=log(x2)^5 ??

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0One more. You're on a roll!

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1lol thanks haha :P but i don't get this one... it looks different.. :/ logx^2...

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1but heres my take on it.. log(x^2)^1 ??

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Take the negative (1) inside, just like the other coefficients ==> exponents.

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1was what i got right?? or no?

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1or is it log(x)^2 ?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0There it is. They are now all connected by addition. Use this guy and bring them all together. log(a) + log(b) = log(a*b)

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1sorry my computer crashed... one sec :/

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1log[(x+4)^−3]+ log(x2)^5+log(x7)^2 + log(x)^2=answer B right? :)

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1use the rule described above to go from log[(x+4)^−3]+ log(x2)^5+log(x7)^2 + log(x)^2 to log[ (x+4)^−3*(x2)^5*(x7)^2*(x)^2 ]

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1kk I'm seeing that :) but how can i simplify it even further?

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1(x+4)^−3 is really 1/[ (x+4)^3 ]

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1same idea applies to (x)^2

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1ok so i get 1/x^2 ?

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1so put that all together

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1so i get answer C right?? :)

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1sorry i meant answer D :) is it answer D then? :)

iheartfood
 2 years ago
Best ResponseYou've already chosen the best response.1@jim_thompson5910 :)
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