If `alpha, beta` are the roots of the equation `ax^(2) + bx + c =0`, prove that
`log_(e) (ax^(2) + bx + c) = log_(e) a + 2 log_(e)x-(1)/(x) (alpha + beta) - (1)/(2x^(2)) (alpha^(2) + beta^(2)) - (1)/(3x^(3)) (alpha^(3) + beta^(3))-...`

Answer

To prove the given equation

\[ \log_e (ax^2 + bx + c) = \log_e a + 2 \log_e x - \frac{1}{x} (\alpha + \beta) - \frac{1}{2x^2} (\alpha^2 + \beta^2) - \frac{1}{3x^3} (\alpha^3 + \beta^3) - \ldots \]

where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(ax^2 + bx + c = 0\), we can follow these steps:

### Step 1: Rewrite the quadratic equation
The quadratic equation can be expressed in terms of its roots:
\[ ax^2 + bx + c = a(x - \alpha)(x - \beta) \]
This is because the roots of the equation are \(\alpha\) and \(\beta\).

### Step 2: Apply the logarithmic property
Using the property of logarithms, we can write:
\[ \log_e (ax^2 + bx + c) = \log_e [a(x - \alpha)(x - \beta)] = \log_e a + \log_e (x - \alpha) + \log_e (x - \beta) \]

### Step 3: Factor out \(x\) from the logarithmic terms
We can factor \(x\) out of the logarithmic terms:
\[ \log_e (x - \alpha) = \log_e x + \log_e \left(1 - \frac{\alpha}{x}\right) \]
\[ \log_e (x - \beta) = \log_e x + \log_e \left(1 - \frac{\beta}{x}\right) \]
Thus, we have:
\[ \log_e (ax^2 + bx + c) = \log_e a + 2 \log_e x + \log_e \left(1 - \frac{\alpha}{x}\right) + \log_e \left(1 - \frac{\beta}{x}\right) \]

### Step 4: Expand the logarithmic terms
Using the Taylor series expansion for \(\log_e(1 - u)\) around \(u = 0\):
\[ \log_e(1 - u) = -u - \frac{u^2}{2} - \frac{u^3}{3} - \ldots \]
we can substitute \(u = \frac{\alpha}{x}\) and \(u = \frac{\beta}{x}\):
\[ \log_e \left(1 - \frac{\alpha}{x}\right) = -\frac{\alpha}{x} - \frac{\alpha^2}{2x^2} - \frac{\alpha^3}{3x^3} - \ldots \]
\[ \log_e \left(1 - \frac{\beta}{x}\right) = -\frac{\beta}{x} - \frac{\beta^2}{2x^2} - \frac{\beta^3}{3x^3} - \ldots \]

### Step 5: Combine the expansions
Combining these expansions gives:
\[ \log_e (ax^2 + bx + c) = \log_e a + 2 \log_e x - \left(\frac{\alpha + \beta}{x} + \frac{\alpha^2 + \beta^2}{2x^2} + \frac{\alpha^3 + \beta^3}{3x^3} + \ldots \right) \]

### Step 6: Identify the final expression
Thus, we have:
\[ \log_e (ax^2 + bx + c) = \log_e a + 2 \log_e x - \frac{1}{x} (\alpha + \beta) - \frac{1}{2x^2} (\alpha^2 + \beta^2) - \frac{1}{3x^3} (\alpha^3 + \beta^3) - \ldots \]
This completes the proof.