If `log_(t)a, a^(t//2) and log_(b)t` are in G.P., then t is equal to

Options

A. `log_(a) (log_(b) a)`
B. `log_(a) (log a) - log_(a) (log b)`
C. `-log_(a) (log_(a)b)`
D. `log_(a) (log b) - log_(a) (log a)`

Answer

To solve the problem where \( \log_t a, a^{(t/2)}, \log_b t \) are in geometric progression (G.P.), we can follow these steps:

### Step 1: Understanding the condition for G.P.
For three terms \( x, y, z \) to be in G.P., the condition is:
\[ y^2 = x \cdot z \]
In our case, let:
- \( x = \log_t a \)
- \( y = a^{(t/2)} \)
- \( z = \log_b t \)

### Step 2: Applying the G.P. condition
Using the G.P. condition:
\[ \left(a^{(t/2)}\right)^2 = \log_t a \cdot \log_b t \]
This simplifies to:
\[ a^t = \log_t a \cdot \log_b t \]

### Step 3: Expressing logarithms in terms of base change
We can express \( \log_t a \) and \( \log_b t \) using the change of base formula:
\[ \log_t a = \frac{\log a}{\log t} \quad \text{and} \quad \log_b t = \frac{\log t}{\log b} \]

### Step 4: Substituting the logarithmic expressions
Substituting these into our equation gives:
\[ a^t = \left(\frac{\log a}{\log t}\right) \cdot \left(\frac{\log t}{\log b}\right) \]
This simplifies to:
\[ a^t = \frac{\log a}{\log b} \]

### Step 5: Rearranging the equation
Rearranging the equation, we have:
\[ a^t \cdot \log b = \log a \]

### Step 6: Taking logarithm on both sides
Taking logarithm on both sides:
\[ \log(a^t \cdot \log b) = \log(\log a) \]
Using the property of logarithms:
\[ t \log a + \log(\log b) = \log(\log a) \]

### Step 7: Isolating \( t \)
Now, isolate \( t \):
\[ t \log a = \log(\log a) - \log(\log b) \]
\[ t = \frac{\log(\log a) - \log(\log b)}{\log a} \]

### Step 8: Final expression for \( t \)
Thus, we have:
\[ t = \frac{\log\left(\frac{\log a}{\log b}\right)}{\log a} \]

### Conclusion
The value of \( t \) is given by:
\[ t = \frac{\log\left(\frac{\log a}{\log b}\right)}{\log a} \]