CAT 2025 Slot 2 QA Question 13: $\log_{64} x^2 + \log_8 \sqrt{y} + 3 \log_{512} (\sqrt{y} z) = 4$, where $x$, $y$ and $z$ are positive real numbers, then the minimum possible value of $(x + y + z)$ is

Question

$\log_{64} x^2 + \log_8 \sqrt{y} + 3 \log_{512} (\sqrt{y} z) = 4$, where $x$, $y$ and $z$ are positive real numbers, then the minimum possible value of $(x + y + z)$ is

Accepted Answer

Convert all logarithms to base $2$.

$\log_{64} x^2 = \dfrac{1}{3} \log_2 x$
$\log_8 \sqrt{y} = \dfrac{1}{6} \log_2 y$
$3 \log_{512} (\sqrt{y} z) = \dfrac{1}{6} \log_2 y + \dfrac{1}{3} \log_2 z$

So the given equation becomes
$\dfrac{1}{3}(\log_2 x + \log_2 y + \log_2 z) = 4$
$\Rightarrow \log_2(xyz

log⁡64x2+log⁡8y+3log⁡512(yz)=4\log_{64} x^2 + \log_8 \sqrt{y} + 3 \log_{512} (\sqrt{y} z) = 4log64​x2+log8​y​+3log512​(y​z)=4, where xxx, yyy and zzz are positive real numbers, then the minimum possible value of (x+y+z)(x + y + z)(x+y+z) is

$\log_{64} x^2 + \log_8 \sqrt{y} + 3 \log_{512} (\sqrt{y} z) = 4$ , where $x$ , $y$ and $z$ are positive real numbers, then the minimum possible value of $(x + y + z)$ is