(iv) f(x)=(x+(1)/(2))log_((x+(1)/(2))){(x^(2)+2x-3)/(4x^(2)-4x-3)}

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(iv)  f(x)=(x+(1)/(2))log_((x+(1)/(2))){(x^(2)+2x-3)/(4x^(2)-4x-3)}

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Identify Base and Argument: Let's first identify the base of the logarithm and the argument of the logarithm in the function f(x). The base of the logarithm is (x + 1/2), and the argument is the fraction (x^2 + 2x - 3)/(4x^2 - 4x - 3).Apply Change of Base Formula: We can apply the change of base formula to the logarithm to convert it to a common logarithm (base 10) or natural logarithm (base e). The change of base formula is log_b(a) = log_c(a) / log_c(b), where c is the new base. However, since the base of the logarithm and the coefficient in front of the logarithm are the same, the logarithm simplifies to 1 when evaluated at the base, which means f(x) simplifies to the argument of the logarithm.Simplify f(x) to Argument: Therefore, we can simplify f(x) to just the argument of the logarithm, which is (x^2 + 2x - 3)/(4x^2 - 4x - 3), since the logarithm of a number at its own base is 1. f(x) = (x + 1/2) * 1 * ((x^2 + 2x - 3)/(4x^2 - 4x - 3))Multiply by Fraction: Now, we can simplify the expression by multiplying (x + 1/2) by the fraction. f(x) = (x + 1/2) * ((x^2 + 2x - 3)/(4x^2 - 4x - 3))Correct Conceptual Error: However, upon reviewing the previous steps, we realize that there has been a conceptual error. The simplification of the logarithm to 1 only applies when the argument of the logarithm is exactly equal to the base, which is not the case here. The argument is a fraction, not just (x + 1/2). Therefore, we cannot simplify the logarithm to 1, and the previous step is incorrect.

(iv) $f(x)=\left(x+\frac{1}{2}\right) \log _{\left(x+\frac{1}{2}\right)}\left\{\frac{x^{2}+2 x-3}{4 x^{2}-4 x-3}\right\}$

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(iv) f(x)=(x+12)log⁡(x+12){x2+2x−34x2−4x−3} f(x)=\left(x+\frac{1}{2}\right) \log _{\left(x+\frac{1}{2}\right)}\left\{\frac{x^{2}+2 x-3}{4 x^{2}-4 x-3}\right\} f(x)=(x+21​)log(x+21​)​{4x2−4x−3x2+2x−3​}

(iv) $f(x)=\left(x+\frac{1}{2}\right) \log _{\left(x+\frac{1}{2}\right)}\left\{\frac{x^{2}+2 x-3}{4 x^{2}-4 x-3}\right\}$