frac{(a - 1) - frac{(a - 1)^2}{2} + frac{(a - | vedclass

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(A) We know that the logarithmic series expansion is given by $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \infty$ for $-1 < x \le 1$. Subs

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$\log_b a$

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(A) We know that the logarithmic series expansion is given by $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \infty$ for $-1 Substituting $x = a - 1$ in the numerator,we get $(a - 1) - \frac{(a - 1)^2}{2} + \frac{(a - 1)^3}{3} - \dots \infty = \ln(1 + a - 1) = \ln(a)$. Similarly,substituting $x = b - 1$ in the denominator,we get $(b - 1) - \frac{(b - 1)^2}{2} + \frac{(b - 1)^3}{3} - \dots \infty = \ln(1 + b - 1) = \ln(b)$. Thus,the expression becomes $\frac{\ln(a)}{\ln(b)}$. Using the change of base formula $\frac{\log_k a}{\log_k b} = \log_b a$,we get $\frac{\ln(a)}{\ln(b)} = \log_b a$.

$\frac{(a - 1) - \frac{(a - 1)^2}{2} + \frac{(a - 1)^3}{3} - \dots \infty}{(b - 1) - \frac{(b - 1)^2}{2} + \frac{(b - 1)^3}{3} - \dots \infty} = $

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