Question: If$\mathbf{2}^{\mathbf{x}}\mathbf{=}\mathbf{4}^{\mathbf{y}}\mathbf{=}\mathbf{8}^{\mathbf{z}}$and\(...

Question

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

Answer 1

Solution

$2^{x} = 2^{2y} = 2^{3z}i.e.,x = 2y = 3z = k$ (say).

Then $xyz = \frac{k^{3}}{6} = 288$ , So $k = 12$

$\therefore x = 12,y = 6,z = 4$ . Therefore, $\frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = \frac{11}{96}$

Question: If\(\mathbf{2}^{\mathbf{x}}\mathbf{=}\mathbf{4}^{\mathbf{y}}\mathbf{=}\mathbf{8}^{\mathbf{z}}\)and\(...

Solution