Question

x²=y⁴=z⁸and xyz=288 then find the value of 1/2y+1/4y+1/8z=?​

Answer 1

It is given that

2^x=4^y=8^z2x=4y=8z

It can be written as

2^x=2^{2y}=2^{3z}2x=22y=23z

x=2y=3zx=2y=3z                   .... (1)

It is also given that

xyz=288xyz=288

(3z)\times (\frac{3z}{2})\times z=288(3z)×(23z)×z=288

(9z^3)=576(9z3)=576

Divide both sides by 9.

z^3=64z3=64

z=4z=4

x=3z=3\times 4=12x=3z=3×4=12

y=\frac{3z}{2}=\frac{3\times 4}{2}=6y=23z=23×4=6

We have to find the value of

\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}2x1+4y1+8z1

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

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

\frac{1}{24}+\frac{1}{24}+\frac{1}{32}241+241+321

\frac{4+4+3}{96}=\frac{11}{96}964+4+3=9611