CAT 2025 Lesson : Logarithm - Applying Log Properties

Which of these equals $\mathrm{log} \ 48?$ (1) $\mathrm{log} \ 4 + \mathrm{log} \ 8$ $\space \space \space \space$
(2) $\mathrm{log} \ 32 + \mathrm{log} \ 16$ $\space \space \space \space$
(3) $\mathrm{log} \ 6 \times \mathrm{log} \ 8$ $\space \space \space \space$
(4) $\mathrm{log} \ 96 - \mathrm{log} \ 2$ $\space \space \space \space$

Solution

Option $(4)$ is correct
$\mathrm{log} \ 96 - \mathrm{log} \ 2$ $= \mathrm{log} \ \dfrac{96}{2} = \mathrm{log} \ 48$

Note that while $\mathrm{log} \ 6 + \mathrm{log} \ 8 = \mathrm{log} \ 48$, option (c) $\mathrm{log} \ 6 \times \mathrm{log} \ 8$ $\ne \mathrm{log} \ 48$

Answer: $(4)$ $\mathrm{log} \ 96 - \mathrm{log} \ 2$

What is the value of $\mathrm{log} (\mathrm{log} \ 100^{500})?$

Solution

With the base not mentioned, this is assumed to be a common logarithm with base $10$.

$\mathrm{log} \ (\mathrm{log} \ 100^{500})$
$= \mathrm{log} \ (\mathrm{log} \ (10^{2})^{500})$ $=\mathrm{log} (\mathrm{log} \ 10^{1000})$
$= \mathrm{log} \ (1000 \mathrm{log} \ 10)$
$= \mathrm{log} \ 1000$
$= \mathrm{log} \ 10^{3} = 3$

Answer: $3$

What is $\mathrm{log} _{\sqrt{32}}{8} = ?$

Solution

$\mathrm{log} _{\sqrt{32}}{8}$ $= 3 \ \mathrm{log} _{\sqrt{32}}{2}$ $= \dfrac{3}{\mathrm{log} _{2}{\sqrt{32}}}$ $= \dfrac{3}{\mathrm{log} _{2}{2^{\frac{5}{2}}}}$ $= \dfrac{3}{\dfrac{5}{2} \times \mathrm{log} _{2}{2}}$ $=\dfrac{6}{5} = 1.2$

Alternatively

$\mathrm{log} _{\sqrt{32}}{8}$ $=\mathrm{log} _{2^{\frac{5}{2}}}{2^{3}}$ $= \dfrac{3}{\frac{5}{2}} \ \mathrm{log} _{2}{2}$ $=\dfrac{6}{5} = 1.2$

Answer: $1.2$

What is $\mathrm{log} _{0.25} {64}$ $=$ ?

Solution

$64 = 2^6$ and $0.5$ $=$ $2^{-2}$ .

So, $\mathrm{log} _{0.25} {64}$ $=$ $\mathrm{log} _{2^{-2}} {2^6}$ $=$ $\dfrac{6}{-2} \times \mathrm{log} _{2} {2}$ $=$ $-3$

Answer: $-3$

Simplify $\mathrm{log} _{\sqrt{y}} {x^2} + \dfrac{1}{\mathrm{log} _{(1/x)} {y^3}}$

Solution

$\mathrm{log} _{\sqrt{y}} {x^2} + \dfrac{1}{\mathrm{log} _{1/x} {y^3}}$ $=$ $\mathrm{log} _{\sqrt{y}} {x^2} + \mathrm{log} _{y^3} {x^{-1}}$

$=$ $\dfrac{2}{0.5}\mathrm{log} _{y} {x} - \dfrac{1}{3} \mathrm{log} _{y} {x}$ $=$ $\mathrm{log} _{y} {x} ( 4 - \dfrac{1}{3})$ $=$ $\dfrac{11}{3} \mathrm{log} _{y} {x}$

Answer: $\dfrac{11}{3} \mathrm{log} _{y} {x}$