If {2^x} = {4^y} = {8^z} and xyz = 288, then | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) ${2^x} = {2^{2y}} = {2^{3z}}\,i.e.,\,x = 2y = 3z = k$ (say).

Then $xyz = {{{k^3}} \over 6} = 288$, So $k = 12$

.$	herefore x = 12,y = 6,z =

Vedclass · Accepted Answer

$11/96$

Vedclass · Answer

(d) ${2^x} = {2^{2y}} = {2^{3z}}\,i.e.,\,x = 2y = 3z = k$ (say).

Then $xyz = {{{k^3}} \over 6} = 288$, So $k = 12$

.$	herefore x = 12,y = 6,z = 4$ Therefore, ${1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = {{11} \over {96}}$

If ${2^x} = {4^y} = {8^z}$ and $xyz = 288,$ then ${1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = $

Similar Questions

The rationalising factor of $2\sqrt 3 - \sqrt 7 $ is

Solution of the equation $\sqrt {(x + 10)} + \sqrt {(x - 2)} = 6$ are

${{\sqrt {6 + 2\sqrt 3 + 2\sqrt 2 + 2\sqrt 6 } - 1} \over {\sqrt {5 + 2\sqrt 6 } }}$

If $x + \sqrt {({x^2} + 1)} = a,$ then $x =$

If ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ then $x =$