Find the value of `1/(log_3e)+1/(log_3e^2)+1/(log_3e^4)+...` is

Options

A. `log_e9`
B. 0
C. 1
D. `log_e270`

Answer

To find the value of the series

\[ S = \frac{1}{\log_3 e} + \frac{1}{\log_3 e^2} + \frac{1}{\log_3 e^4} + \ldots \]

we can start by rewriting the terms in a more manageable form using the properties of logarithms.

### Step 1: Rewrite the terms using logarithm properties

Using the property of logarithms, we have:

\[ \log_3 e^n = n \cdot \log_3 e \]

Thus, we can rewrite the series as:

\[ S = \frac{1}{\log_3 e} + \frac{1}{2 \log_3 e} + \frac{1}{4 \log_3 e} + \ldots \]

### Step 2: Factor out the common term

Notice that each term has a common factor of \(\log_3 e\):

\[ S = \frac{1}{\log_3 e} \left(1 + \frac{1}{2} + \frac{1}{4} + \ldots\right) \]

### Step 3: Identify the series inside the parentheses

The series inside the parentheses is a geometric series where the first term \(a = 1\) and the common ratio \(r = \frac{1}{2}\).

### Step 4: Find the sum of the geometric series

The sum \(S_g\) of an infinite geometric series can be calculated using the formula:

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

Substituting the values:

\[ S_g = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

### Step 5: Substitute back to find \(S\)

Now substituting back into our expression for \(S\):

\[ S = \frac{1}{\log_3 e} \cdot 2 = \frac{2}{\log_3 e} \]

### Step 6: Use the change of base formula

Using the change of base formula for logarithms, we can express \(\log_3 e\) in terms of natural logarithms:

\[ \log_3 e = \frac{\log_e e}{\log_e 3} = \frac{1}{\log_e 3} \]

Thus, we have:

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

### Step 7: Final result

The final result can be expressed as:

\[ S = \log_e 9 \]

### Conclusion

Thus, the value of the series is:

\[ \log_e 9 \]