log question...
- Thread starter k9fireman
- Start date
pka
Elite Member
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- Jan 29, 2005
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\(\displaystyle \L
\begin{array}{l}
x\log (2) = (x + 1)\log (3) \\
x\log (2) = x\log (3) + \log (3) \\
x\log (2) - x\log (3) = \log (3) \\
x(\log (2) - \log (3)) = \log (3) \\
x = \frac{{\log (3)}}{{(\log (2) - \log (3))}} = \frac{{\log (3)}}{{\log (2/3)}} \\
\end{array}\)
\begin{array}{l}
x\log (2) = (x + 1)\log (3) \\
x\log (2) = x\log (3) + \log (3) \\
x\log (2) - x\log (3) = \log (3) \\
x(\log (2) - \log (3)) = \log (3) \\
x = \frac{{\log (3)}}{{(\log (2) - \log (3))}} = \frac{{\log (3)}}{{\log (2/3)}} \\
\end{array}\)