Question

If $x^{\left(y^z\right)}=2^8$ then the maximum possible value of the product $x.y.z$ where $x,y,z$ positive integers is

• A. $16$
• B. $12$
• C. $256$
• D. $24$

Answer 1

Answer: (A)

Given equation is,

$x^{y^z}=2^8$

$\Rightarrow x=2$ [Since, equated the bases]

$\therefore x=2$..........(1)

$\Rightarrow y^z=8$ [Since, equated the powers]

there is only one possibility, where $y^z=8$

therefore, we have,

$y^z=8=2^3$ [Since, $2^3=2\times 2\times 2=8$]

$\therefore y=2$...........(2)

$\therefore z=3$...........(3)

Using the value (1), (2) and (3), we get the value of $x\times y\times x$ as,

$x\times y\times z=2\times 2\times 3=12$

$12>0$

Therefore, the maximum possible value of $x\times y\times z$ where $x,y,z>0$ is $12$.

Hence, Option C is correct.