CAT 2025 Slot 2 QA Question & Solution
AlgebraMedium
Question
If $\log_{64}{x^{2}+\log_{8}{\sqrt{y}+3\log_{512}{(\sqrt{y}z)}}}=4$, where $x$, $y$ and $z$ are positive real numbers, then the minimum possible value of $(x+y+z)$ is
Options
48
36
24
96
Solution
$$64 = 8^2 \text{and} 512 = 8^3$$
$$\log_{64}{x^{2}+\log_{8}{\sqrt{y}+3\log_{512}{(\sqrt{y}z)}}}=4$$,
Property of log: $$\log_{b^m}\ a^{n\ }=\frac{n}{m}\ \log_ba$$
$$\log_{8^2}{x^{2}+\log_{8}{\sqrt{y}+3\log_{8^3}{(\sqrt{y}z)}}}=4$$
Using the above-mentioned property, the expression becomes $$\log_{8}{x}+\log_{8}{\sqrt{y}+\log_{8}{(\sqrt{y}z)}}=4$$
$$\log_8x\sqrt{y}\cdot(\sqrt{y}z)=4$$
$$\log_8xyz=4$$
$$xyz =8^4=2^{12}$$
Using AM-GM inequality
$$\frac{\left(x+y+z\right)}{3}\ge\sqrt[\ 3]{xyz}$$
$$\frac{\left(x+y+z\right)}{3}\ge2^4$$
$$\left(x+y+z\right)\ge48$$
