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  4. The value of 7 log((16/15) ) +5 log((25/24)) + 3 log((81/80)) is equla to

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Q. The value of $7 \log\left(\frac{16}{15} \right) +5 \log\left(\frac{25}{24}\right) + 3 \log\left(\frac{81}{80}\right) $ is equla to

BITSATBITSAT 2011

Solution:

$\therefore 7 \log \left(\frac{16}{15}\right)+5 \log \left(\frac{25}{24}\right)+3 \log \left(\frac{81}{80}\right)$
$=\log \left[\left(\frac{16}{15}\right)^{7} \cdot\left(\frac{25}{24}\right)^{5} \cdot\left(\frac{81}{80}\right)^{3}\right]$
$=\log 2$