If log 2=x, log 3=y and log 7=z, then the val | vedclass

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(B) Given: $\log 2=x, \log 3=y, \log 7=z$.We need to find the value of $\log (4 \sqrt[3]{63})$.$\log (4 \sqrt[3]{63}) = \log (2^2 \cdot (3^2 \cdot 7)^

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$2 x+\frac{2}{3} y+\frac{1}{3} z$

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(B) Given: $\log 2=x, \log 3=y, \log 7=z$.We need to find the value of $\log (4 \sqrt[3]{63})$.$\log (4 \sqrt[3]{63}) = \log (2^2 \cdot (3^2 \cdot 7)^{1/3})$Using the logarithmic properties $\log (ab) = \log a + \log b$ and $\log (a^n) = n \log a$:$\log (4 \sqrt[3]{63}) = \log (2^2) + \log ((3^2 \cdot 7)^{1/3})$$= 2 \log 2 + \frac{1}{3} \log (3^2 \cdot 7)$$= 2 \log 2 + \frac{1}{3} (\log 3^2 + \log 7)$$= 2 \log 2 + \frac{1}{3} (2 \log 3 + \log 7)$Substituting the given values:$= 2x + \frac{1}{3} (2y + z)$$= 2x + \frac{2}{3} y + \frac{1}{3} z$.

If $\log 2=x, \log 3=y$ and $\log 7=z,$ then the value of $\log (4 \sqrt[3]{63})$ is

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