Question

2^x=4^y=8^z and xyz=288 then find 1/2x +1/4y+1/8z

Answer 1

Answer:

The correct option is D.

Step-by-step explanation:

It is given that

$2^x=4^y=8^z$

It can be written as

$2^x=2^{2y}=2^{3z}$

$x=2y=3z$                   .... (1)

It is also given that

$xyz=288$

$(3z)\times (\frac{3z}{2})\times z=288$

$(9z^3)=576$

Divide both sides by 9.

$z^3=64$

$z=4$

$x=3z=3\times 4=12$

$y=\frac{3z}{2}=\frac{3\times 4}{2}=6$

We have to find the value of

$\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}$

Substitute x=12, y=6 adn z=4.

$\frac{1}{2(12)}+\frac{1}{4(6)}+\frac{1}{8(4)}$

$\frac{1}{24}+\frac{1}{24}+\frac{1}{32}$

$\frac{4+4+3}{96}=\frac{11}{96}$

Therefore the correct option is D.

Answer 2

"Answer: 8

Given       ${ 2 }^{ x }\quad =\quad { 4 }^{ y }\quad =\quad { 8 }^{ z }{ 2 }^{ x }\quad =\quad { 2 }^{ (2y) }\quad =\quad { 2 }^{ (3z) }$

If basics are equal then their powers are also equal

       x = 2y = 3z

xyz=288

$x\times (\frac { x }{ 2 } )\times (\frac { x }{ 3 } )\quad =\quad 238 (x^3)/6 = 288{ x }^{ 3 }\quad =\quad 288\times 6\quad =\quad 1728$

       x= 12

$y\quad =\quad \frac { x }{ 2 } \quad =\quad \frac { 12 }{ 2 } \quad =\quad 6$

$z\quad =\quad \frac { x }{ 3 } \quad =\quad \frac { 12 }{ 3 } \quad =\quad 4$

$\frac { 1 }{ 2 } x+\frac { 1 }{ 4 } y+\frac { 1 }{ 8 } z\quad =\quad (\frac { 1 }{ 2 } )\times 12+(\frac { 1 }{ 4 } )\times 6+(\frac { 1 }{ 8 } )\times 4\quad =\quad 6+(\frac { 3 }{ 2 } )+(\frac { 1 }{ 2 } )\quad =\quad 8$"