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    5^3w-1 = 4^250
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    (Original post by SmartFailure)
    5^3w-1 = 4^250
    Hm yes, this question makes sense. Before we do your homework for you answer me this: what have you tried and also, is the question this:
    5^(3w-1) = 4^250
    or this:
    (5^3w)-1 = 4^250
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    5^(3w-1) = 4^250

    I've tried doing:

    log(5)^3w-1 = log(4)^250
    log (5)^3w-1 - log (4)^250 = 0
    log (5^3w-1 / 4^250) = 0

    Don't know where to go from here
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    (Original post by SmartFailure)
    5^(3w-1) = 4^250

    I've tried doing:

    log(5)^3w-1 = log(4)^250
    log (5)^3w-1 - log (4)^250 = 0
    log (5^3w-1 / 4^250) = 0

    Don't know where to go from here
    Go back to step one. \log(5^{3w-1})=(3w-1)\log(5)
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    is this right:

    (3w-1)log(5) = 250log(4)

    3w-1 = 250xlog(4) / log(5)

    3w - 1 = 215.33
    3w = 216.33
    w = 72.11
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    (Original post by SmartFailure)
    is this right:

    (3w-1)log(5) = 250log(4)

    3w-1 = 250xlog(4) / log(5)

    3w - 1 = 215.33
    3w = 216.33
    w = 72.11
    Give the exact form, unless it asks for the approximation. Overall, yes that's right.
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    The only way to calculate this is: 250xlog(4) / log(5) is by calculator right?
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    (Original post by SmartFailure)
    The only way to calculate this is: 250xlog(4) / log(5) is by calculator right?
    No need, just leave it in its exact form.
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    Alright thanks for the help
 
 
 
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