# The number of cells in a tumor doubles every 3.5 months. If the tumor begins with a single cell,...

## Question:

The number of cells in a tumor doubles every 3.5 months. If the tumor begins with a single cell, how many cells will there be after?

A. 44 years

B. 66 years

## Exponential Growth Equations:

An exponential growth explains a situation whereby something grows in relation to its present value. When something grows exponentially, its growth increases with time. An exponential growth equation assumes the form {eq}P_t = P_0e^{rt} {/eq}.

The exponential growth equation takes the form:

\begin{align} P_t = P_0e^{rt} \end{align}

where:

{eq}P_0 {/eq} is the initial population,

{eq}P_t {/eq} is the population at time {eq}t {/eq},

{eq}r {/eq} is the interest rate and,

{eq}t {/eq} is the time.

In our question, we have:

{eq}\begin{align} & P_0 = 1\\[0.3cm] & P_{3.5} = 2\\[0.3cm] & t = 3.5 \end{align} {/eq}

Therefore:

\begin{align} & 2= 1e^{3.5r}\\[0.3cm] \end{align}

Introducing natural logarithms on both sides:

\begin{align} & \ln 2= \ln e^{3.5r}\\[0.3cm] & \ln 2= 3.5r \ln e\\[0.3cm] & r = \dfrac{\ln 2}{3.5} \\[0.3cm] & r \approx 0.553295 \end{align}

Therefore, the growth equation for the cells is:

\begin{align} P_t = e^{0.553295t} \end{align}

Using the equation above, the number of cells when {eq}t = 44\times 12 = 528 {/eq} months is:

\begin{align} & P_{44} = e^{0.553295\times 528}\\[0.3cm] & \boxed{\color{blue}{P_{44} \approx 2.585\times 10^{45}}} \; \rm cells \end{align}.

The number of cells when {eq}t = 66\times 12 = 792 {/eq} months is:

\begin{align} & P_{66} = e^{0.553295\times 792}\\[0.3cm] & \boxed{\color{blue}{P_{66} \approx 1.315\times 10^{68} }} \; \rm cells \end{align} 