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    Hi Guys,

    Could you please help me answer this question:

    a^0.31 = (ab)^c

    express c in terms of b?

    Any help is greatly appreciated,
    Thanks in advance
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    (Original post by KishoreV)
    Hi Guys,

    Could you please help me answer this question:

    a^0.31 = (ab)^c

    express c in terms of b?

    Any help is greatly appreciated,
    Thanks in advance
    What chapter of maths is this question in? Because this will normally give you an idea of how to solve it. For example, I straight away thought to use logs, but the title says in indices, and in edexcel maths, these are two different chapters?
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    (Original post by SheldonWannabe)
    What chapter of maths is this question in? Because this will normally give you an idea of how to solve it. For example, I straight away thought to use logs, but the title says in indices, and in edexcel maths, these are two different chapters?
    Hi Sheldon, sorry this is not in any books. Its one of the problems I'm trying to solve my self to understand indices better.

    Could you please give how you would answer this using logs?
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    (Original post by KishoreV)
    Hi Sheldon, sorry this is not in any books. Its one of the problems I'm trying to solve my self to understand indices better.

    Could you please give how you would answer this using logs?
    Where is this question from? And are you sure this is solvable just using the laws of indices?
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    (Original post by KishoreV)
    Hi Sheldon, sorry this is not in any books. Its one of the problems I'm trying to solve my self to understand indices better.

    Could you please give how you would answer this using logs?
    I don't think you can rearrange it to find c using only laws of indices, but this is how to do it using the laws of logs. (if you're not familiar with log laws it might be a bit strange)

    a^0.31 = (ab)^c
    0.31loga = cloga + clogb
    c(loga + logb) = 0.31loga
    c = (0.31loga)/(loga + logb)

    Knowing me, there's probably a silly error in their some where. But I can't think of any other way to solve it anyway!
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    (Original post by SheldonWannabe)
    I don't think you can rearrange it to find c using only laws of indices, but this is how to do it using the laws of logs. (if you're not familiar with log laws it might be a bit strange)

    a^0.31 = (ab)^c
    0.31loga = cloga + clogb
    c(loga + logb) = 0.31loga
    c = (0.31loga)/(loga + logb)

    Knowing me, there's probably a silly error in their some where. But I can't think of any other way to solve it anyway!
    But you've still got 'a' in your final line, don't you need to get rid of this so that c is expressed in terms of b?
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    (Original post by KishoreV)
    Hi Guys,

    Could you please help me answer this question:

    a^0.31 = (ab)^c

    express c in terms of b?

    Any help is greatly appreciated,
    Thanks in advance

    Are a and b co-prime
    is there any other information
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    (Original post by Username_valid)
    But you've still got 'a' in your final line, don't you need to get rid of this so that c is expressed in terms of b?
    haha! sorry, I thought it said in terms of a AND b, I'm such an idiot, I thought it seemed to easy. :dry:
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    Either the question is wrong or info is missing. c must depend on a.
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    (Original post by SheldonWannabe)
    haha! sorry, I thought it said in terms of a AND b, I'm such an idiot, I thought it seemed to easy. :dry:
    Aha no worries, I've made worse mistakes. I don't think that question could be answered anyway unless I'm missing something haha
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    (Original post by KishoreV)
    Hi Sheldon, sorry this is not in any books. Its one of the problems I'm trying to solve my self to understand indices better.

    Could you please give how you would answer this using logs?
    Is this a "real" question, or is it one you have made up yourself to try to test your understanding? If it's the latter, then you may have invented something that can't be simplified in the way you want. If it's the former, then as others have pointed out, there needs to be more information supplied to arrive at a solution.
 
 
 
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