Question

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Explanation:

Part (a)

Step1: Recall the division of functions

The division of two functions \( (f \div g)(x) \) is defined as \( \frac{f(x)}{g(x)} \) where \( g(x)
eq 0 \).
Given \( f(x)=2^{x}-1 \) and \( g(x)=\log(x - 1) \), then \( (f\div g)(x)=\frac{2^{x}-1}{\log(x - 1)} \)

Step2: Ensure the denominator is non - zero

We know that for the function \( \log(x - 1) \), the argument \( x-1>0\) (domain of logarithm function) and also \( \log(x - 1)
eq0 \). But for the expression \( \frac{f(x)}{g(x)} \), the primary definition from the operation of function division is \( \frac{f(x)}{g(x)}=\frac{2^{x}-1}{\log(x - 1)} \) with the condition that \( g(x)
eq0 \) and \( g(x) \) is defined.

Step1: Analyze the domain of \( f(x) \) and \( g(x) \)

• For \( f(x)=2^{x}-1 \), the domain of the exponential function \( 2^{x} \) is all real numbers \( (-\infty,\infty) \), so the domain of \( f(x) \) is \( (-\infty,\infty) \).
• For \( g(x)=\log(x - 1) \), the domain of a logarithmic function \( \log(u) \) requires \( u>0 \). So for \( g(x)=\log(x - 1) \), we need \( x - 1>0\Rightarrow x>1 \). Also, for the function \( (f\div g)(x)=\frac{f(x)}{g(x)} \), the denominator \( g(x)=\log(x - 1)

eq0 \).

• Solve \( \log(x - 1)=0 \). Since \( \log_{a}1 = 0 \) (for \( a>0,a

eq1 \)), if we assume the logarithm is base 10 (common logarithm) or base \( e \) (natural logarithm), \( \log(x - 1)=0\Rightarrow x - 1 = 1\Rightarrow x=2 \).

Step2: Combine the conditions

The domain of \( (f\div g)(x) \) must satisfy:

• \( x-1>0 \) (from the domain of the logarithmic function in \( g(x) \))
• \( \log(x - 1)

eq0 \) (from the denominator not being zero)

From \( x - 1>0 \), we have \( x>1 \). From \( \log(x - 1)
eq0 \), we have \( x-1
eq1\Rightarrow x
eq2 \).

So the domain is \( (1,2)\cup(2,\infty) \)

Answer:

\( (f\div g)(x)=\frac{2^{x}-1}{\log(x - 1)} \)

Part (b)