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Question : 30 of 86

Marks: +1, -0

If

a^{x}=b^{y}=c^{z}

b^{2}=a c,

y

equals:

$\;\frac{x y}{x+z}$
$\;\frac{z x}{2(x-z)}$
$\;\frac{z x}{2(z-x)}$
$\;\frac{2 x z}{x+z}$

Solution:

Concept:

Use the common value of powers to express

a

,

b

,

c

in terms of

x

,

y

,

z

and then apply

b^2 = ac

.

Explanation:

Let

a^x = b^y = c^z = k

. Then

a = k^{1/x}

,

b = k^{1/y}

,

c = k^{1/z}

.

Substitute into

b^2 = ac

:

(k^{1/y})^2 = k^{1/x} \cdot k^{1/z}

⟹

k^{2/y} = k^{1/x + 1/z}

.

Equating exponents:

\frac{2}{y} = \frac{1}{x} + \frac{1}{z} = \frac{x+z}{xz}

.

Thus

y = \frac{2xz}{x+z}

.

Answer:

y = \frac{2xz}{x+z}

Option D:

\frac{2 x z}{x+z}

is correct.

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