If `f(x) = sin^(-1) (sin ("log"_(2) x))`, then find the value of `f(300)`

Explanation

To solve the problem, we need to find the value of \( f(300) \) where \( f(x) = \sin^{-1}(\sin(\log_2 x)) \).

### Step 1: Substitute \( x = 300 \)
We start by substituting \( x \) with \( 300 \) in the function:
\[ f(300) = \sin^{-1}(\sin(\log_2 300)) \]

### Step 2: Calculate \( \log_2 300 \)
Next, we need to calculate \( \log_2 300 \). We can use the change of base formula:
\[ \log_2 300 = \frac{\log_{10} 300}{\log_{10} 2} \]
Using approximate values, we find:
\[ \log_{10} 300 \approx 2.4771 \quad \text{and} \quad \log_{10} 2 \approx 0.3010 \]
Thus,
\[ \log_2 300 \approx \frac{2.4771}{0.3010} \approx 8.22 \]

### Step 3: Determine the range of \( \log_2 300 \)
Since \( \log_2 256 = 8 \) and \( \log_2 512 = 9 \), we have:
\[ 8 < \log_2 300 < 9 \]

### Step 4: Use the property of \( \sin^{-1}(\sin x) \)
The function \( \sin^{-1}(\sin x) \) returns \( x \) if \( x \) is within the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). However, since \( 8.22 \) is outside this range, we need to adjust it.

### Step 5: Adjust \( \log_2 300 \) to fit within the range
To adjust \( \log_2 300 \) into the range of \( \sin^{-1} \), we can use the periodic property of the sine function:
\[ \sin(x) = \sin(\pi - x) \]
Thus, we can write:
\[ \sin(\log_2 300) = \sin(\pi - \log_2 300) \]
This gives us:
\[ f(300) = \sin^{-1}(\sin(\pi - \log_2 300)) = \pi - \log_2 300 \]

### Step 6: Final calculation
Now we can substitute back the value of \( \log_2 300 \):
\[ f(300) = \pi - 8.22 \]

### Conclusion
Thus, the value of \( f(300) \) is:
\[ f(300) \approx \pi - 8.22 \]