`32^(tan^2x)+32^(Sec^2x)=81`. Find number of solution if `0 le x le pi/4`.

Answer

To solve the equation \( 32^{\tan^2 x} + 32^{\sec^2 x} = 81 \) for the interval \( 0 \leq x \leq \frac{\pi}{4} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 32^{\tan^2 x} + 32^{\sec^2 x} = 81 \]
Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can rewrite \( 32^{\sec^2 x} \):
\[ 32^{\sec^2 x} = 32^{1 + \tan^2 x} = 32 \cdot 32^{\tan^2 x} \]
Substituting this back into the equation gives:
\[ 32^{\tan^2 x} + 32 \cdot 32^{\tan^2 x} = 81 \]

### Step 2: Factor the equation
We can factor out \( 32^{\tan^2 x} \):
\[ 32^{\tan^2 x} (1 + 32) = 81 \]
This simplifies to:
\[ 32^{\tan^2 x} \cdot 33 = 81 \]

### Step 3: Isolate \( 32^{\tan^2 x} \)
Now, we isolate \( 32^{\tan^2 x} \):
\[ 32^{\tan^2 x} = \frac{81}{33} \]
Calculating the right-hand side:
\[ 32^{\tan^2 x} = \frac{81}{33} \approx 2.4545 \]

### Step 4: Take the logarithm
Next, we take the logarithm (base 32) of both sides:
\[ \tan^2 x = \log_{32}\left(\frac{81}{33}\right) \]
Using the change of base formula, we can express this as:
\[ \tan^2 x = \frac{\log\left(\frac{81}{33}\right)}{\log(32)} \]

### Step 5: Calculate the value
Calculating the logarithm:
\[ \log_{32}\left(\frac{81}{33}\right) \approx 0.25855 \]
Thus, we have:
\[ \tan^2 x \approx 0.25855 \]

### Step 6: Find \( \tan x \)
Taking the square root:
\[ \tan x \approx \sqrt{0.25855} \approx 0.5085 \]

### Step 7: Determine the solutions
Now, we need to find the number of solutions for \( \tan x = 0.5085 \) in the interval \( 0 \leq x \leq \frac{\pi}{4} \). The function \( \tan x \) is increasing in this interval.

### Conclusion
Since \( \tan x \) is a continuous and increasing function in the interval \( [0, \frac{\pi}{4}] \), there will be exactly one solution for \( \tan x = 0.5085 \).

Thus, the number of solutions is:
\[ \boxed{1} \]