Q. 21 - Calculus

Question

Use the new definition of $\ln x$ from Definition $4.36$ to argue that

1. $\ln x$ has domain $\left(0,\infty \right)$ and range $\mathrm{ℝ}$.
2. $e^x$ has domain $\mathrm{ℝ}$ and range $\left(0,\infty \right)$.

Step-by-Step Solution

Answer

Part $\left(a\right)$: $\ln x$ has domain $\left(0,\infty \right)$ and range $\mathrm{ℝ}$.

Part $\left(b\right)$: $e^x$ has domain $\mathrm{ℝ}$ and range $\left(0,\infty \right)$.

Part a Step 1 . Given information

$\ln x$ has domain $\left(0,\infty \right)$ and range $\mathrm{ℝ}$.

Part a Step 2 . Consider the following figure of ln   x .

Part a Step 3 . In the above figure, it is seen that the function ln   x is defined on 0 , ∞ as it is continuous on 0 , ∞ .

So, the domain of $\ln x$ is $\left(0,\infty \right)$.

Now, $\ln x$ is $0$ for $x=1$.It increases without bound as $x\to {\infty }^{+}$ and decreases without bound as $x\to 0^{+}$.

So, the range of $\ln x$ is, $\left(-\infty ,\infty \right)$.

Therefore, $\ln x$ has domain $\left(0,\infty \right)$ and range $\mathrm{ℝ}$.

Part b Step 1 . The objective is to argue that e x has domain ℝ and range 0 , ∞ .

Now, 

$e^x=\frac{1}{\ln x}$

Hence, the domain for $\ln x$ becomes the range for $e^x$ and the range of $\ln x$ becomes the domain for $e^x$.

Therefore, $e^x$ has domain $\mathrm{ℝ}$ and range $\left(0,\infty \right)$.