91Ó°ÊÓ

To explain why \(x^{t}=e^{t \ln x}\) for every number \(t\), we will use the properties of logarithms and exponents. Recall that the natural logarithm function, denoted by \(\ln\), has the property that \(e^{\ln x} = x\). Using this property, we can rewrite the expression as: \(x^{t} = e^{\ln(x^{t})}\) Now, let's apply the power rule for logarithms, which states that \(\ln(x^{t}) = t \ln x\). So, we have: \(x^{t} = e^{t \ln x}\) This is the equality we set out to prove.

To explain why \(\frac{x^{t}-1}{t} \approx \ln x\) if \(t\) is close to 0, we will use the definition of the derivative. Consider the function \(f(t) = x^{t}\). We want to find the derivative of \(f(t)\) with respect to \(t\). By the definition of a derivative, we have: \(f'(t) = \lim_{t \to 0} \frac{f(t) - f(0)}{t} = \lim_{t \to 0} \frac{x^{t} - x^{0}}{t}\) Since \(x^0 = 1\) for any positive number \(x\), we can simplify this expression to: \(f'(t) = \lim_{t \to 0} \frac{x^{t} - 1}{t}\)

Using the equality from part (a), we know that \(x^t = e^{t \ln x}\). So, we have: \(f'(t) = \lim_{t \to 0} \frac{e^{t \ln x} - 1}{t}\) Now, we apply the chain rule, which tells us that if we have a function \(g(t) = e^{h(t)}\), then \(g'(t) = e^{h(t)}h'(t)\). In our case, \(g(t) = x^{t}\) and \(h(t) = t \ln x\). Hence, we have \(g'(t) = e^{t \ln x}\cdot (\ln x)\) or: \(f'(t) = e^{t \ln x}\cdot \ln x\) Plugging in \(t = 0\): \(f'(0) = e^{0 \ln x}\cdot \ln x\) Since \(e^0 = 1\), our expression simplifies to: \(f'(0) = 1 \cdot \ln x = \ln x\) Since we derived that \(f'(t) = \lim_{t \to 0} \frac{x^{t} - 1}{t}\), when \(t\) is close to \(0\), we have: \(\frac{x^{t} - 1}{t} \approx \ln x\) This is the relationship we aimed to explain in part (b).