The vlaue of `9^(1//3)xx9^(1//9)xx9^(1//27)xx………oo` is :

Options

A. 9
B. 3
C. 1
D. None of theses

Explanation

To solve the problem, we need to evaluate the infinite product:

\[ 9^{\frac{1}{3}} \times 9^{\frac{1}{9}} \times 9^{\frac{1}{27}} \times \ldots \]

### Step 1: Combine the Exponents
Using the property of exponents that states \( a^m \times a^n = a^{m+n} \), we can combine the exponents:

\[ 9^{\left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\right)} \]

### Step 2: Identify the Series
The series in the exponent is:

\[ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \]

This is a geometric series where the first term \( a = \frac{1}{3} \) and the common ratio \( r = \frac{1}{3} \).

### Step 3: Sum the Infinite Geometric Series
The sum \( S \) of an infinite geometric series can be calculated using the formula:

\[ S = \frac{a}{1 - r} \]

Substituting the values of \( a \) and \( r \):

\[ S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

### Step 4: Substitute Back into the Exponent
Now we substitute the sum back into the exponent:

\[ 9^{\frac{1}{2}} \]

### Step 5: Simplify the Expression
We know that \( 9^{\frac{1}{2}} \) is the square root of 9:

\[ 9^{\frac{1}{2}} = \sqrt{9} = 3 \]

### Final Answer
Thus, the value of the infinite product is:

\[ \boxed{3} \]