If `log_(e )k=2[(3)/(5)+(1)/(3)((3)/(5))^(3)+(1)/(5)((3)/(5))^(5)+….oo]` then k =

Options

A. 2
B. 3
C. 4
D. 8

Answer

To solve the given problem, we start with the equation:

\[ \log_e k = 2\left[\frac{3}{5} + \frac{1}{3}\left(\frac{3}{5}\right)^3 + \frac{1}{5}\left(\frac{3}{5}\right)^5 + \ldots \right] \]

### Step 1: Identify the series
The series inside the brackets can be rewritten as:

\[ S = \frac{3}{5} + \frac{1}{3}\left(\frac{3}{5}\right)^3 + \frac{1}{5}\left(\frac{3}{5}\right)^5 + \ldots \]

### Step 2: Recognize the pattern
Notice that the series can be expressed in terms of a power series. The general term appears to be:

\[ \frac{1}{n} \left(\frac{3}{5}\right)^{2n-1} \quad \text{for } n = 1, 2, 3, \ldots \]

### Step 3: Factor out the common term
We can factor out the common term \(2\):

\[ S = 2\left[\frac{3}{5} + \frac{1}{3}\left(\frac{3}{5}\right)^3 + \frac{1}{5}\left(\frac{3}{5}\right)^5 + \ldots \right] \]

### Step 4: Rewrite the series
The series can be rewritten using the formula for the sum of an infinite geometric series. The series can be expressed as:

\[ S = \frac{3/5}{1 - (3/5)^2} \]

### Step 5: Substitute \(x = \frac{3}{5}\)
Using the formula for the sum of a geometric series, we substitute \(x = \frac{3}{5}\):

\[ S = \frac{\frac{3}{5}}{1 - \frac{9}{25}} = \frac{\frac{3}{5}}{\frac{16}{25}} = \frac{3}{5} \cdot \frac{25}{16} = \frac{15}{16} \]

### Step 6: Substitute back into the logarithm
Now, substituting back into our logarithmic equation:

\[ \log_e k = 2 \cdot \frac{15}{16} \]

### Step 7: Simplify the logarithm
This simplifies to:

\[ \log_e k = \frac{30}{16} = \frac{15}{8} \]

### Step 8: Solve for \(k\)
To find \(k\), we exponentiate both sides:

\[ k = e^{\frac{15}{8}} \]

### Final Answer
Thus, the value of \(k\) is:

\[ k = e^{\frac{15}{8}} \]