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Math Help - logarithms question ♥

  1. #1
    Member grgrsanjay's Avatar
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    logarithms question ♥

    a,b,c are positive integers forming an increasing G.P and b-a is a perfect cube and log_6(a)+ log_6(b)+ log_6(c) =6 then a+b+c=?
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  2. #2
    Senior Member eumyang's Avatar
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    \begin{aligned}<br />
\log_6\,a + \log_6\,b + \log_6\,c &= 6 \\<br />
6^{\log_6\,a + \log_6\,b + \log_6\,c} &= 6^6 \\<br />
6^{\log_6\,a} \cdot 6^{\log_6\,b} \cdot 6^{\log_6\,c} &= 6^6 \\<br />
abc &= 6^6<br />
\end{aligned}

    If a, b, and c form a geometric progression then b = ar and c = ar^2 (for some ratio r):
    \begin{aligned}<br />
abc &= 6^6 \\<br />
a(ar)(ar^2) &= 6^6 \\<br />
a^3 r^3 &= 6^6 \\<br />
ar &= 36 \\<br />
b &= 36<br />
\end{aligned}

    b - a has to be a perfect cube, or 36 - a has to be a perfect cube. There are a few possibilities for a (35, 28, 9), but the only one that will work in our case is if a = 9.

    If a = 9 and b = 36, then r = 4, which makes c = 144. So a + b + c = 189.
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