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Math Help - Power rule

  1. #1
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    Power rule

    If a and b are nonzero real numbers and (3.92)a = (7.86)b, what is the value of (b/a)^2? The answer is .44. The answer says use logs, so I took the log of both sides. Then they say to use the power rule. I thought the a and b would have to be an exponent to use the power rule? Do they, or is the answer explanation I'm reading correct?
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  2. #2
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    Re: Power rule

    Quote Originally Posted by benny92000 View Post
    If a and b are nonzero real numbers and (3.92)a = (7.86)b, what is the value of (b/a)^2? The answer is .44. The answer says use logs, so I took the log of both sides. Then they say to use the power rule. I thought the a and b would have to be an exponent to use the power rule? Do they, or is the answer explanation I'm reading correct?
    (3.92)a = (7.86)b

    => b/a = 3.92/7.86

    => (b/a)^2 = 0.2487.

    I hate to break it to you sport, but the given answer for the question as posted is wrong (and it does not require logs).

    Perhaps you need to go back and post the correct question ....?
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  3. #3
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    Re: Power rule

    I think that answers the question. It is a typo (from the part of the book that I'm using). I assume a and b should be exponents.
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  4. #4
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    Re: Power rule

    Is there a reason why you chose NOT to tell us that the a and b were exponents to start with? That changes the entire problem.

    If (3.92)^a = (7.86)^b, then, yes, use logarithms. log(3.92^a)= alog(3.92)= log(7.86^b)= blog(7.86) so that
    \frac{b}{a}= \frac{log(3.92)}{log(7.86)}
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  5. #5
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    Re: Power rule

    It was a typo in the question. Thanks though for the help.
    Last edited by mr fantastic; September 18th 2011 at 08:35 PM.
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