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If we have an algebraic equation as $a^{x} = b^{y} = c^{z}$ and $b^{2} = ac$ , then value of y is equal to :

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\frac{2 xz}{x + z}

\frac{xz}{2 (x - z)}

\frac{xz}{2 (z - x)}

\frac{2 xz}{x - z}

answer is A.

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Detailed Solution

Concept- Firstly, use the above equation to the value of a and b in terms of c. And then substitute the values of a and b in terms of c.
As given,

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

From this we can say that,

a^{x} = c^{z}

and

b^{y} = c^{z}

Using the identity,

k^{m} = n \Rightarrow k = n^{\frac{1}{m}}

, we get

a = c^{\frac{z}{x}}

_______ (i)
Similarly for

b^{y} = c^{z}

We get,

b = c^{\frac{z}{y}}

_______ (ii)
Now if we substitute the values of a and b from eq(i) & eq(ii)
in the given equation ,

b^{2} = ac

we get,

c^{\frac{2 z}{y}} = c^{\frac{z}{x}} . c

Now, as we know

k^{l} . k^{m} = k^{l + m}

c^{\frac{2 z}{y}} = c^{\frac{z}{x} + 1}

Taking log on both the sides and use the identity,

\log \log k^{m} = m \log log k

\log \log c^{\frac{2 z}{y}} = \log \log c^{\frac{z}{x} + 1}

\Rightarrow \frac{2 z}{y} \log log c = (\frac{z}{x} + 1) \log log c

______(iii)
Canceling log from both sides-

\Rightarrow \frac{2 z}{y} = (\frac{z}{x} + 1)

\Rightarrow \frac{2 z}{y} = (\frac{z + x}{z})

Taking reciprocal on both sides-

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

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

Hence, the correct answer is option 1.

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