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Question : 27 of 86

Marks: +1, -0

If

2^{x}=4^{y}=8^{z}

\left(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}\right)=\frac{24}{7}

, then the value of

z

$\frac{7}{16}$
$\frac{7}{32}$
$\frac{7}{48}$
$\frac{7}{64}$

Solution:

Concept:

Express all terms with a common base (2) to relate

x

,

y

,

z

.

Explanation:

Let

2^{x}=4^{y}=8^{z}=k

.

Then

x=\log_2 k

,

y=\log_4 k=\frac{1}{2}\log_2 k

,

z=\log_8 k=\frac{1}{3}\log_2 k

.

Set

t=\log_2 k

, so

x=t

,

y=\frac{t}{2}

,

z=\frac{t}{3}

.

Substitute into

\frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}=\frac{24}{7}

:

\frac{1}{2t}+\frac{1}{4\cdot\frac{t}{2}}+\frac{1}{6\cdot\frac{t}{3}} = \frac{1}{2t}+\frac{1}{2t}+\frac{1}{2t}=\frac{3}{2t}=\frac{24}{7}

.

Cross-multiply:

3\cdot7=24\cdot2t \Rightarrow 21=48t \Rightarrow t=\frac{21}{48}=\frac{7}{16}

.

Then

z=\frac{t}{3}=\frac{7}{16}\cdot\frac{1}{3}=\frac{7}{48}

.

Answer:

z=\frac{7}{48}

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