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    mickey97's Avatar
    mickey97 Posts: 1, Reputation: 1
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    #1

    Feb 6, 2010, 01:35 PM
    solve y = 6^(x+1) - 6 for x without logs in the expression
    For example, solving the equasion y = 2^(x-4) for x yields x = y^(1/2) + 4. But I can't seem to do the same for y = 6^(x+1) - 6
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Feb 8, 2010, 08:36 AM

    Your example is wrong. Consider if x = 5; then y = 2^(x-4)= 2^1 = 2. But your "solution" says x = y^(1/2) + 4 = sqrt(2) + 4, which is certainly not equal to 5. Your error here is in thinking that the square root of 2^(x-4) is x-4.

    Why your aversion to using logs?
    harum's Avatar
    harum Posts: 339, Reputation: 27
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    #3

    Feb 9, 2010, 09:21 PM

    I doubt you can avoid logarithms in your problem. A neat trick might help you calculate logarithms faster, if this is why you choose to avoid them:

    (log(base b) a) = (log(base c) a) / (log(base c) b);

    For example, (log(base 6) 10) = ln(10) / ln(6), or = lg(10)/lg(6).
    "ln" is log(base e), "lg" is log(base 10).
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #4

    Feb 10, 2010, 05:22 AM





    Factor:



    Now, continue and use logs.

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