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anonymous
 one year ago
please help, question below
anonymous
 one year ago
please help, question below

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mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0You need to know these rules of logs: \(\Large \log_b \dfrac{x}{y} = \log_bx  \log_b y\) \(\Large \log_b x^n = n \log_b x\) Use the first rule first. What do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that means that in the second part of the equation, it always has to be subtracting... okay. that takes 2 answers off my list.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it be \[\log_{4}82\log_{4}x \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0someone please help me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did i get the right answer??:D

radar
 one year ago
Best ResponseYou've already chosen the best response.0No, that would be dividing 8 squared by c

radar
 one year ago
Best ResponseYou've already chosen the best response.0Look at the earlier post by mathstudent55, you are supposed to be subtracting the divisor.

radar
 one year ago
Best ResponseYou've already chosen the best response.0Did you choose D "did i get the right answer??:D"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, thats a super smiley face, haha. i choose C and it was correct

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0Applying the first rule we get: \( \Large \log_4 \dfrac{8}{x^2} = \log_4 8  \log_4 x^2\) Applying the second rule we get: \( \Large \log_4 \dfrac{8}{x^2} = \log_4 8  2 \log_4 x\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0You are correct. Good job!

radar
 one year ago
Best ResponseYou've already chosen the best response.0Sorry about that I guess I am not "hip" to internet speak lol.

radar
 one year ago
Best ResponseYou've already chosen the best response.0I thought you were on track that is why I said you are doing fine, then you posted that and it threw me off.

radar
 one year ago
Best ResponseYou've already chosen the best response.0@mathstudent55 Just a problem in "internet slang"
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