Question:
\({\log _{12}}81 = p,then\;3\left( {\frac{{4 - p}}{{4 + p}}} \right)\) is equal to
- log416
- log68
- log616
- log28
Correct Answer: 2
$\log _{12} 81=\mathrm{p} \Rightarrow \log _{12} 3^{4}=\mathrm{p}$
$\Rightarrow 4 \log _{12} 3=\mathrm{p}$
$\Rightarrow \frac{\mathrm{p}}{4}=\log _{12} 3$
$3\left(\frac{4-\mathrm{p}}{4+\mathrm{p}}\right)=3\left(\frac{1-\frac{\mathrm{p}}{4}}{1+\frac{\mathrm{p}}{4}}\right)$
$=3\left(\frac{1-\log _{12} 3}{1+\log _{12} 3}\right)$
$=3\left(\frac{\log _{12} 12-\log _{12} 3}{\log _{12} 12+\log _{12} 3}\right)$
$=3\left(\frac{\log (12 / 3)}{\log (12 / 3)}\right)$
$=3 \frac{\log 4}{\log 36}=3 \log _{36} 4$
$=\log _{6} 8$
$\log _{12} 81=\mathrm{p} \Rightarrow \log _{12} 3^{4}=\mathrm{p}$
$\Rightarrow 4 \log _{12} 3=\mathrm{p}$
$\Rightarrow \frac{\mathrm{p}}{4}=\log _{12} 3$
$3\left(\frac{4-\mathrm{p}}{4+\mathrm{p}}\right)=3\left(\frac{1-\frac{\mathrm{p}}{4}}{1+\frac{\mathrm{p}}{4}}\right)$
$=3\left(\frac{1-\log _{12} 3}{1+\log _{12} 3}\right)$
$=3\left(\frac{\log _{12} 12-\log _{12} 3}{\log _{12} 12+\log _{12} 3}\right)$
$=3\left(\frac{\log (12 / 3)}{\log (12 / 3)}\right)$
$=3 \frac{\log 4}{\log 36}=3 \log _{36} 4$
$=\log _{6} 8$
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