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  4. The value of (64) ((1/ log 9 16))+( log 5 4 ⋅ log 6 5 ⋅ log 7 6)- log 7 28 is equal to

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Q. The value of (64) ${ }^{\left(\frac{1}{\log _9 16}\right)}+\left(\log _5 4 \cdot \log _6 5 \cdot \log _7 6\right)-\log _7 28$ is equal to

Continuity and Differentiability

Solution:

$ (64)^{\log _{16} 9}+\log _7 4-\log _7 28 $
$=4^{3 \log _4 3}+\log _7\left(\frac{4}{28}\right)$
$=27-1=26$