Solve for a,`lamda` if `log_lamda acdotlog_(5)lamdacdotlog_(lamda)25=2`.

Answer

To solve the equation \( \log_{\lambda} a \cdot \log_{5} \lambda \cdot \log_{\lambda} 25 = 2 \), we can follow these steps:

### Step 1: Write the equation
We start with the given equation:
\[ \log_{\lambda} a \cdot \log_{5} \lambda \cdot \log_{\lambda} 25 = 2 \]

### Step 2: Use the change of base formula
Using the change of base formula, we can express the logarithms in terms of natural logarithms (or any common base):
\[ \log_{\lambda} a = \frac{\log a}{\log \lambda}, \quad \log_{5} \lambda = \frac{\log \lambda}{\log 5}, \quad \log_{\lambda} 25 = \frac{\log 25}{\log \lambda} \]
Substituting these into the equation gives:
\[ \frac{\log a}{\log \lambda} \cdot \frac{\log \lambda}{\log 5} \cdot \frac{\log 25}{\log \lambda} = 2 \]

### Step 3: Simplify the equation
Notice that \( \log \lambda \) cancels out:
\[ \frac{\log a \cdot \log 25}{\log 5 \cdot \log \lambda} = 2 \]
Now, we can express \( \log 25 \) as \( \log 5^2 = 2 \log 5 \):
\[ \frac{\log a \cdot 2 \log 5}{\log 5 \cdot \log \lambda} = 2 \]

### Step 4: Cancel \( \log 5 \)
The \( \log 5 \) terms cancel out:
\[ \frac{2 \log a}{\log \lambda} = 2 \]

### Step 5: Simplify further
Dividing both sides by 2 gives:
\[ \frac{\log a}{\log \lambda} = 1 \]

### Step 6: Cross-multiply
Cross-multiplying yields:
\[ \log a = \log \lambda \]

### Step 7: Take antilogarithm
Taking the antilogarithm of both sides results in:
\[ a = \lambda \]

### Final Answer
Thus, the solution is:
\[ a = \lambda \]