If `(1)/(log_(a)x) + (1)/(log_(b)x) = (2)/(log_(c)x)`, prove that : `c^(2) = ab`.

Answer

To prove that \( c^2 = ab \) given the equation

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

we can follow these steps:

### Step 1: Rewrite the logarithms using the change of base formula.
Using the change of base formula, we can express the logarithms as:

\[ \log_a x = \frac{\log x}{\log a}, \quad \log_b x = \frac{\log x}{\log b}, \quad \text{and} \quad \log_c x = \frac{\log x}{\log c}. \]

### Step 2: Substitute the rewritten logarithms into the equation.
Substituting these into the original equation gives:

\[ \frac{1}{\frac{\log x}{\log a}} + \frac{1}{\frac{\log x}{\log b}} = \frac{2}{\frac{\log x}{\log c}}. \]

This simplifies to:

\[ \frac{\log a}{\log x} + \frac{\log b}{\log x} = \frac{2 \log c}{\log x}. \]

### Step 3: Combine the left-hand side.
Combining the left-hand side:

\[ \frac{\log a + \log b}{\log x} = \frac{2 \log c}{\log x}. \]

### Step 4: Eliminate the common denominator.
Since \(\log x\) is common in both sides and assuming \(\log x \neq 0\), we can eliminate it:

\[ \log a + \log b = 2 \log c. \]

### Step 5: Use the properties of logarithms.
Using the property of logarithms that states \(\log m + \log n = \log(mn)\), we can rewrite the left-hand side:

\[ \log(ab) = 2 \log c. \]

### Step 6: Rewrite the right-hand side.
Using the property that \(k \log m = \log(m^k)\), we can rewrite the right-hand side:

\[ \log(ab) = \log(c^2). \]

### Step 7: Equate the arguments of the logarithms.
Since the logarithms are equal, we can equate their arguments:

\[ ab = c^2. \]

### Conclusion
Thus, we have proved that

\[ c^2 = ab. \]