Question

If 2 4 8 x y z
  and xyz  288 then value of 1 1 1
2 4 8 x y z
  is:

Answer 1

Step-by-step explanation:

It is given that

2^x=4^y=8^z2

x

=4

y

=8

z

It can be written as

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

x

=2

2y

=2

3z

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

It is also given that

xyz=288xyz=288

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

2

3z

)×z=288

(9z^3)=576(9z

3

)=576

Divide both sides by 9.

z^3=64z

3

=64

z=4z=4

x=3z=3\times 4=12x=3z=3×4=12

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

2

3z

=

2

3×4

=6

We have to find the value of

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

2x

1

+

4y

1

+

8z

1

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}

24

1

+

24

1

+

32

1

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

96

4+4+3

=

96

11