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Math Help - Rewrite Log Function in terms of A and B?

  1. #1
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    Unhappy Rewrite Log Function in terms of A and B?

    This has been driving me up the wall, I have a test the day after tomorrow and this is DEFINITELY going to be on the test. It seems simple, but for some reason I can't get the hang of it, please help.

    ***Rewrite the equation in terms of A and B

    A=Log34 , B=log35

    this equation--> log3(15/2)


    I've tried 3b+(a/2), but thats wrong. I know why it's wrong, but I dont know how to get the right answer.
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  2. #2
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    Re: Rewrite Log Function in terms of A and B?

    Quote Originally Posted by chopper07 View Post
    This has been driving me up the wall, I have a test the day after tomorrow and this is DEFINITELY going to be on the test. It seems simple, but for some reason I can't get the hang of it, please help.

    ***Rewrite the equation in terms of A and B

    A=Log34 , B=log35

    this equation--> log3(15/2)


    I've tried 3b+(a/2), but thats wrong. I know why it's wrong, but I dont know how to get the right answer.
    Hint: \log_3(15)=1+B
    Thanks from chopper07
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: Rewrite Log Function in terms of A and B?

    A = \log_3(4) = \log_3(2^2) = 2 log_3(2), so  \log_3(2) = \frac A 2.

     \log_3(15/2) = \log_3(\frac {3*5} 2 ) = \log_3(3) + \log_3 (5) - \log_3 (2)

    You have values for \log_3(5) and \log_3 (2). The only thing left is  \log_3(3). Recall that if \log_a b = c then  a^c = b. Here a = 3 and b = 3, so what's the value of c? Hence what's the value of  \log_3 (3)? Post back with your solution and we'll check it for you.
    Last edited by ebaines; October 8th 2012 at 02:31 PM.
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    Re: Rewrite Log Function in terms of A and B?

    Ohh, ok i get it now. thank you !!
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  5. #5
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    Re: Rewrite Log Function in terms of A and B?

    Wait, so would it be 1+B+(a/2) ?
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