If x is a positive real number different from 1 such that `log_a x, log_b x, log_c x` are in A.P then

Options

A. `logb=2((loga)(logc))/((loga+logc))`
B. `b=(a+c)/2`
C. `b=sqrtac`
D. `c^2=(ac)^(log_ab`

Answer

To solve the problem where \( \log_a x, \log_b x, \log_c x \) are in Arithmetic Progression (A.P.), we can follow these steps:

### Step 1: Understanding the condition for A.P.
If three numbers \( a, b, c \) are in A.P., then the middle term is equal to the average of the first and third terms. Thus, we can write:

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

### Step 2: Rewrite the logarithms using the change of base formula
Using the change of base formula, we can rewrite the logarithms:

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

Substituting these into the A.P. condition gives:

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

### Step 3: Simplifying the equation
Multiplying both sides by \( 2 \log b \):

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

### Step 4: Factor out \( \log x \)
Assuming \( \log x \neq 0 \) (since \( x \) is a positive real number different from 1), we can divide both sides by \( \log x \):

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

### Step 5: Finding a common denominator
Rearranging gives:

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

### Step 6: Cross-multiplying
Cross-multiplying gives:

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

### Step 7: Rearranging the equation
Rearranging the equation yields:

\[ 2 \log a \cdot \log c - \log c = \log b \]

### Step 8: Factoring out \( \log c \)
Factoring out \( \log c \):

\[ \log c (2 \log a - 1) = \log b \]

### Step 9: Exponentiating to remove the logarithm
Exponentiating both sides gives:

\[ c^{2 \log a - 1} = b \]

### Step 10: Final expression
This can be rearranged to:

\[ c^2 = ac^{\log_b a} \]

### Conclusion
Thus, we conclude that:

\[ c^2 = ac^{\log_b a} \]