Chapter 3.2, Problem 15E

Solution

To determine: the exponential function that satisfies the given condition initial mass $=0.6\text{g}$ and doubling every $3$ day.

The exponential function is $0.6{\left(2\right)}^{\frac{x}{3}}$ .

Given information:

The initial mass $m_0$ is $0.6\text{g}$ .

The mass is doubling every $3$ day.

Concept Used:

If a population $P$ is changing at a constant percentage rate $r$ each year, then $P\left(t\right)=P_0{\left(1+r\right)}^t$ where $P_0$ is the initial population, $r$ is expressed as a decimal, and $t$ is time in years.

If $r>0$ , then $P\left(t\right)$ is an exponential growth function, and its growth factor is the base of the exponential function, $1+r$ .

Calculation:

Let $m\left(x\right)$ be the mass that changes at a constant growth rate $r$ per day, then the mass growth function is $m\left(x\right)=m_0{\left(1+r\right)}^x$ .

Here, $m_0=0.6$

  $r=2\left(\text{doubled}\right)$

Doubling every $3$ days means $\frac{x}{3}$ because $x$ is time in years.

Substitute $0.6$ for $m_0$ , $2$ for $r$ and replace $x$ with $\frac{x}{3}$ in $m\left(x\right)=m_0{\left(1+r\right)}^x$ .

  $m\left(x\right)=0.6{\left(2\right)}^{\frac{x}{3}}$

Hence, the exponential function is $0.6{\left(2\right)}^{\frac{x}{3}}$ .