1) If log 2 $=x$, log 3 $=y$ and log 7 $=z$, then the value of log $(4.\sqrt[3]{69})$ is : A) $2x+\frac{2}{3}y$ $-\frac{1}{3}z$ B) $2x+\frac{2}{3}y$ $+\frac{1}{3}z$ C) $2x-\frac{2}{3}y$ $+\frac{1}{3}z$ D) $-2x+\frac{2}{3}y$ $+\frac{1}{3}z$ Answer: Option BExplanation:$log (4.\sqrt[3]{63})$ = log 4 + log $(\sqrt[3]{63})$ = log 4 + log $(63)^{1/3}$ = log $(2^{2})$ + log $(7\times 3^{2})^{1/3}$ = 2log 2 $+\frac{1}{3}$ log 7$+\frac{2}{3}$ log 3 $=2x+\frac{1}{3}z$ $+\frac{2}{3}y$
