2^x=4^y=8^z an xyz=288 then 1/2x+1/4y+1/8z
Answers
Answer:
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
2^x = 4^y = 8^z , xyz = 288
2^x = 4^y
2^x = 2^(2y) here base is same. So powers are also same.
so, x =2y (1)
also , 4^y = 8^z
2^(2y) = 2^3z
2y = 3z
z=2y/3 ( 2)
xyz =288
2y*y*2y/3 = 288
4 y^3 /3 = 288
4 y^3 = 3*288
y^3 =3*288/4 = 216 =6^3
so, y=6
x=2y
x = 2*6 =12
z= 2y/3
z = 2*6/3 =4
1/2x = 1/2*12 = 1/24
1/4y = 1/4*6 = 1/24
1/8z = 1/8*4 = 1/32
1/2x +1/4y +1/8z = 1/24 +1/24 +1/32
=(4+4+3)/96
= 11/96