# The rate of auto thefts triples every 8 months. (a) Determine, to two decimal places, the base b...

## Question:

The rate of auto thefts triples every 8 months.

(a) Determine, to two decimal places, the base {eq}b {/eq} for an exponential model {eq}y = Ab^t {/eq} of the rate of auto thefts as a function of time in months.

(b) Find the doubling time to the nearest tenth of a month.

## Exponential Model

Many quantities can be represented with an exponential function. Some key words to look for that may indicate that a quantity increases or decreases exponentially are "doubling time" or "half-life." In that case, we can model the quantity using the function {eq}y=Ab^t {/eq} for some values of {eq}A {/eq} and {eq}b {/eq}.

a) Let {eq}T(t) {/eq} be the rate of auto thefts at time {eq}t {/eq}, measured in months. Then we're given that {eq}T(t) {/eq} is of the form {eq}T(t)=Ab^t {/eq}. Since the number of thefts triples every 8 months, we also have {eq}T(8)=3T(0) {/eq}.

Note that {eq}T(0)=Ab^0=A {/eq}, so {eq}T(8)=3A {/eq}.

{eq}\begin{align*} 3A&=T(8)\\ &=Ab^8\\ 3&=b^8&&\text{(dividing both sides by }A\text{)}\\ 3^{1/8}&=b&&\text{(since the base of an exponential function must be positive).} \end{align*} {/eq}

So the base of the exponential model must be {eq}b=3^{1/8} {/eq}. To two decimal places, we have {eq}\boxed{b \approx 1.15}\, {/eq}.

b) The doubling time will be the time {eq}t {/eq} such that {eq}T(t)=2T(0)=2A {/eq}. We can solve this equation for {eq}t {/eq} as follows:

{eq}\begin{align*} 2A&=T(t)\\ &=Ab^t\\ 2&=b^t\\ \ln 2 &= t \ln b&&\text{(taking the logarithm of both sides)}\\ &= t \ln 3^{1/8}&&\text{(by the value of }b\text{ found in part a))}\\ &=t\left(\frac{\ln 3}{8}\right)\\ \frac{8 \ln 2}{\ln 3}&=t &&\text{(isolating }t\text{).} \end{align*} {/eq}

So the doubling time is {eq}t=\frac{8\ln 2}{\ln 3} {/eq}. To the nearest tenth of a month, the doubling time is therefore {eq}\boxed{t \approx 4.8\text{ months}}\, {/eq}.