If `(1)/(log_(a) x) + (1)/(log_(c) x) = (2)/(log_(b)x ) ` then a, b, c, are in

Options

A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer

To solve the equation

\[ \frac{1}{\log_a x} + \frac{1}{\log_c x} = \frac{2}{\log_b x} \]

we will use the change of base formula for logarithms, which states that

\[ \log_a x = \frac{\log x}{\log a} \]

Using this, we can rewrite the equation:

1. Rewrite each term using the change of base formula:
\[ \frac{1}{\log_a x} = \frac{\log a}{\log x}, \quad \frac{1}{\log_c x} = \frac{\log c}{\log x}, \quad \text{and} \quad \frac{2}{\log_b x} = \frac{2 \log b}{\log x} \]

2. Substitute these into the original equation:
\[ \frac{\log a}{\log x} + \frac{\log c}{\log x} = \frac{2 \log b}{\log x} \]

3. Since all terms have a common denominator of \(\log x\), we can multiply through by \(\log x\) (assuming \(\log x \neq 0\)):
\[ \log a + \log c = 2 \log b \]

4. Using the properties of logarithms, we can combine the left-hand side:
\[ \log(ac) = 2 \log b \]

5. Rewrite the right-hand side:
\[ \log(ac) = \log(b^2) \]

6. Since the logarithms are equal, we can equate the arguments:
\[ ac = b^2 \]

This shows that \(a\), \(b\), and \(c\) are in a geometric progression (G.P.) because in a G.P., the square of the middle term is equal to the product of the other two terms.

### Final Answer:
Thus, \(a\), \(b\), and \(c\) are in geometric progression.

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