If x is numerically so small so that `x^2` and higher powers of x can be neglected , then `(1+(2x)/(3))^(3/2), (32 + 5x)^(-1/5)` is approximately equal to :

Options

A. `(32 + 31x)/(64)`
B. `(32 +32x)/(64)`
C. `(31 + 32x)/(64)`
D. `(1-2x)/(64)`

Answer

To solve the problem, we need to evaluate the expression \( (1 + \frac{2x}{3})^{\frac{3}{2}} \cdot (32 + 5x)^{-\frac{1}{5}} \) while neglecting \( x^2 \) and higher powers of \( x \).

### Step-by-Step Solution:

1. **Evaluate \( (1 + \frac{2x}{3})^{\frac{3}{2}} \)**:
Using the binomial approximation for small \( x \), we have:
\[ (1 + x)^n \approx 1 + nx \]
Here, \( n = \frac{3}{2} \) and \( x = \frac{2x}{3} \):
\[ (1 + \frac{2x}{3})^{\frac{3}{2}} \approx 1 + \frac{3}{2} \cdot \frac{2x}{3} = 1 + x \]

2. **Evaluate \( (32 + 5x)^{-\frac{1}{5}} \)**:
First, factor out 32:
\[ (32 + 5x)^{-\frac{1}{5}} = 32^{-\frac{1}{5}} \cdot \left(1 + \frac{5x}{32}\right)^{-\frac{1}{5}} \]
Now, using the binomial approximation again:
\[ (1 + u)^{-m} \approx 1 - mu \quad \text{for small } u \]
Here, \( m = \frac{1}{5} \) and \( u = \frac{5x}{32} \):
\[ (1 + \frac{5x}{32})^{-\frac{1}{5}} \approx 1 - \frac{1}{5} \cdot \frac{5x}{32} = 1 - \frac{x}{32} \]
Thus,
\[ (32 + 5x)^{-\frac{1}{5}} \approx \frac{1}{32} \cdot \left(1 - \frac{x}{32}\right) = \frac{1}{32} - \frac{x}{1024} \]

3. **Combine the two results**:
Now, we multiply the two approximations:
\[ (1 + x) \cdot \left(\frac{1}{32} - \frac{x}{1024}\right) \]
Expanding this gives:
\[ = \frac{1}{32} + x - \frac{x}{1024} \]
Neglecting \( x^2 \) and higher powers, we can simplify this to:
\[ = \frac{1}{32} + x \left(1 - \frac{1}{1024}\right) \approx \frac{1}{32} + x \]

4. **Final Result**:
Therefore, the expression \( (1 + \frac{2x}{3})^{\frac{3}{2}} \cdot (32 + 5x)^{-\frac{1}{5}} \) is approximately equal to:
\[ \frac{1}{32} + x \]