If `a != 1 and l n a^(2) + (l n a^(2))^(2) + (l n a^(2))^(3) + ... = 3 (l n a + (l n a)^(2) + ( ln a)^(3) + (l n a)^(4) + ....)` then 'a' is equal to

Options

A. `e^(1//5)`
B. `sqrte`
C. `root(3)(e)`
D. `root(4)(e)`

Answer

To solve the equation given in the problem, we start by analyzing the left-hand side (LHS) and right-hand side (RHS) of the equation separately.

### Step 1: Identify the series on both sides
The LHS is:
\[ \ln(a^2) + (\ln(a^2))^2 + (\ln(a^2))^3 + \ldots \]
This is an infinite geometric series where the first term \(a_1 = \ln(a^2)\) and the common ratio \(r = \ln(a^2)\).

The RHS is:
\[ 3 \left( \ln(a) + (\ln(a))^2 + (\ln(a))^3 + \ldots \right) \]
This is also an infinite geometric series where the first term \(b_1 = \ln(a)\) and the common ratio \(r = \ln(a)\).

### Step 2: Calculate the sum of the series
The sum of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a_1}{1 - r} \]
For the LHS:
\[ S_{LHS} = \frac{\ln(a^2)}{1 - \ln(a^2)} = \frac{2\ln(a)}{1 - \ln(a^2)} \]
For the RHS:
\[ S_{RHS} = 3 \cdot \frac{\ln(a)}{1 - \ln(a)} \]

### Step 3: Set the LHS equal to the RHS
Now we equate the two sums:
\[ \frac{2\ln(a)}{1 - \ln(a^2)} = 3 \cdot \frac{\ln(a)}{1 - \ln(a)} \]

### Step 4: Simplify the equation
We can simplify \(1 - \ln(a^2)\) as follows:
\[ 1 - \ln(a^2) = 1 - 2\ln(a) \]
Thus, the equation becomes:
\[ \frac{2\ln(a)}{1 - 2\ln(a)} = \frac{3\ln(a)}{1 - \ln(a)} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 2\ln(a)(1 - \ln(a)) = 3\ln(a)(1 - 2\ln(a)) \]

### Step 6: Expand both sides
Expanding both sides results in:
\[ 2\ln(a) - 2\ln^2(a) = 3\ln(a) - 6\ln^2(a) \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ -2\ln^2(a) + 6\ln^2(a) + 2\ln(a) - 3\ln(a) = 0 \]
This simplifies to:
\[ 4\ln^2(a) - \ln(a) = 0 \]

### Step 8: Factor the equation
Factoring out \(\ln(a)\):
\[ \ln(a)(4\ln(a) - 1) = 0 \]

### Step 9: Solve for \(\ln(a)\)
This gives us two cases:
1. \(\ln(a) = 0\) which implies \(a = 1\) (not valid since \(a \neq 1\)).
2. \(4\ln(a) - 1 = 0\) which implies \(\ln(a) = \frac{1}{4}\).

### Step 10: Find the value of \(a\)
Taking the antilogarithm:
\[ a = e^{\frac{1}{4}} \]

Thus, the final value of \(a\) is:
\[ \boxed{e^{\frac{1}{4}}} \]