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    Could I have some assistance to do this problem ?
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    (Original post by JackSpinner1)
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    Could I have some assistance to do this problem ?
    Have you made any progress? Post all your working / ideas.

    If you don't know where to start then I would begin by getting rid of the logs to make things easier e.g.

    \log_b a = 2 \Rightarrow a = b^2
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    I ended up with a=b^2, c-3 = b^3 and c+5 = a^2. from these equations I arrived c-3 = a(a) and so c-3 and c+5 = a^2. Would that show that this is contradictory?
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    (Original post by JackSpinner1)
    I ended up with a=b^2, c-3 = b^3 and c+5 = a^2. from these equations I arrived c-3 = a(a) and so c-3 and c+5 = a^2. Would that show that this is contradictory?
    c-3 = a(a)

    How did you get that?
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    My bad, I see what i did wrong there
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    Could i have help of where to go from here, and what does the 'specifies a uniquely' mean as one of the solutions.
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    (Original post by JackSpinner1)
    Could i have help of where to go from here, and what does the 'specifies a uniquely' mean as one of the solutions.
    That means that there is a single value of 'a' that satisfies these equations.

    So you have a=b^2, c-3 = b^3 and c+5 = a^2

    It shouldn't be too hard to eliminate a and c from these equations and leave an equation only in terms of b. Have you tried that? Post all your algebra if you get stuck.
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    I ended up with c+5 = b^4 by subbing in a=b^2 into the c+5 equation.
    I then subbed in c=b^3+3 into c+5=b^4 to get b^4-b^3=8. Solved for b i got 2 and the other solution has to be negative so i disregarded it, so i got b =2 c= 11 and a = 4.
    And so A is the solution?
 
 
 
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