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Surds and Indices Practice Test 2

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Question : 16 of 17

Marks: +1, -0

Given that

2^x = 4^y = 8^z

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

, then find the value of z.

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

Solution:

Concept:

If powers of the same base are equal, then their exponents are equal.

Explanation:

Given

2^x = 4^y = 8^z

.

Write bases as powers of

2

:

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

.

So,

x = 2y = 3z

.

Thus

x = 3z

and

y = \frac{3z}{2}

.

Substitute in

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

.

\frac{1}{2(3z)}+\frac{1}{4(\frac{3z}{2})}+\frac{1}{6z}=\frac{24}{7}

.

Simplify:

\frac{1}{6z}+\frac{1}{6z}+\frac{1}{6z}=\frac{24}{7}

.

\frac{3}{6z}=\frac{24}{7}

.

\frac{1}{2z}=\frac{24}{7}

.

2z=\frac{7}{24}

.

z=\frac{7}{48}

.

Answer:

z = \frac{7}{48}

.

So the correct option is B.

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