Logarithms | Arithmetic aptitude | Discussion Page

If log 2 $=x$ , log 3 $=y$ and log 7 $=z$ , then the value of log $(4.\sqrt[3]{69})$ is :

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

$-\frac{1}{3}z$

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

$+\frac{1}{3}z$

$2x-\frac{2}{3}y$

$+\frac{1}{3}z$

$-2x+\frac{2}{3}y$

$+\frac{1}{3}z$

Option B

$log (4.\sqrt[3]{63})$ = 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

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

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