Find the range of x for which the binomial expansions of the following are valid .
`(7 + 3x)^(-5)`

Answer

To find the range of \( x \) for which the binomial expansion of \( (7 + 3x)^{-5} \) is valid, we can follow these steps:

### Step 1: Identify the form of the binomial expansion
The binomial expansion is valid for expressions of the form \( (1 + u)^n \) where \( n \) is not a natural number, provided that \( |u| < 1 \).

### Step 2: Rewrite the expression
We can rewrite \( (7 + 3x)^{-5} \) in a suitable form. We factor out \( 7 \) from the expression:

\[ (7 + 3x)^{-5} = 7^{-5} \left(1 + \frac{3x}{7}\right)^{-5} \]

### Step 3: Identify \( u \) in the expression
In this case, \( u = \frac{3x}{7} \). For the binomial expansion to be valid, we need:

\[ |u| < 1 \]

### Step 4: Set up the inequality
Substituting \( u \):

\[ \left| \frac{3x}{7} \right| < 1 \]

### Step 5: Solve the inequality
This inequality can be solved as follows:

1. Multiply both sides by \( 7 \) (which is positive, so the inequality sign remains the same):

\[ |3x| < 7 \]

2. Divide both sides by \( 3 \):

\[ |x| < \frac{7}{3} \]

### Step 6: Express the solution in interval notation
The inequality \( |x| < \frac{7}{3} \) implies:

\[ -\frac{7}{3} < x < \frac{7}{3} \]

Thus, the range of \( x \) for which the binomial expansion is valid is:

\[ x \in \left(-\frac{7}{3}, \frac{7}{3}\right) \]

### Final Answer
The range of \( x \) is \( x \in \left(-\frac{7}{3}, \frac{7}{3}\right) \).

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