`((y-1)-(1)/(2)(y-1)^(2)+(1)/(3)(y-1)^(3)-....oo)/((a-1)-(1)/(2)(a-1)^(2)+(1)/(3)(a-1)^(3)-....oo)=`

Options

A. `log_(y)a`
B. `log_(a)y`
C. `log_(a)(y//a)`
D. `log_(e )(a//y)`

Answer

To solve the given expression

\[ \frac{(y-1) - \frac{1}{2}(y-1)^2 + \frac{1}{3}(y-1)^3 - \ldots}{(a-1) - \frac{1}{2}(a-1)^2 + \frac{1}{3}(a-1)^3 - \ldots} \]

we can recognize that both the numerator and the denominator can be expressed in terms of logarithms.

### Step 1: Identify the series as a logarithm
The series in the numerator resembles the Taylor series expansion for \(\log(1+x)\), which is:

\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

In our case, we can set \(x = (y-1)\) for the numerator and \(x = (a-1)\) for the denominator.

### Step 2: Write the logarithmic expressions
Thus, we can express the numerator and denominator as:

\[ \text{Numerator: } \log(y) \quad \text{(since } y-1 = x \text{)} \]
\[ \text{Denominator: } \log(a) \quad \text{(since } a-1 = x \text{)} \]

### Step 3: Substitute back into the expression
Now substituting these back into the original expression gives us:

\[ \frac{\log(y)}{\log(a)} \]

### Step 4: Apply the change of base formula
Using the change of base formula for logarithms, we have:

\[ \frac{\log(y)}{\log(a)} = \log_a(y) \]

### Final Result
Thus, the final result of the given expression is:

\[ \log_a(y) \]