ABC26GN2054 · Logarithms
Subject: General Aptitude · Chapter: Logarithms · Exam: · Marks: · Difficulty:
If $\log 2=x, \log 3=y$ and $\log 7=z$, then the value of $\log (4 \cdot \sqrt[3]{63})$ is
(a)$2 x+\frac{2}{3} y-\frac{1}{3} z$
(b)$2 x+\frac{2}{3} y+\frac{1}{3} z$
(c)$2 x-\frac{2}{3} y+\frac{1}{3} z$
(d)$-2 x+\frac{2}{3} y+\frac{1}{3} z$
Answer
Explanation
$\log (4 \cdot \sqrt[3]{63})=\log 4+\log (\sqrt[3]{63})=\log 4+\log (63)^{1 / 3}=\log$ $$\begin{array}{r} \left(2^{2}\right)+\log \left(7 \times 3^{2}\right)^{1 / 3} \\ =2 \log 2+\frac{1}{3} \log 7+\frac{2}{3} \log 3=2 x+\frac{1}{3} z+\frac{2}{3} y . \end{array}$$
Explanation as extracted from the printed page; notation may be imperfect.
Open in whiteboard · Browse this chapter in the app