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  4. The value of log √2( operatornameantilog128((3/7))) equals

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Q. The value of $\log _{\sqrt{2}}\left(\operatorname{antilog}_{128}\left(\frac{3}{7}\right)\right)$ equals

Continuity and Differentiability

Solution:

antilog $\log _{128}\left(\frac{3}{7}\right)=(128)^{\frac{3}{7}}=\left(2^7\right)^{\frac{3}{7}}=2^3$.
$\therefore \log _{\sqrt{2}}(2)^3=\frac{3}{1 / 2} \log _2 2=6$