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$\left(0,\infty \right)$.

Short 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$\left(0,\infty \right)$.

Step-by-step solution

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 role="math" localid="1648817423905" $0$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 ex 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$\left(0,\infty \right)$.