Question

Devise the exponential growth function that fits the given data; then answer the accompanying questions. Be sure to identify the reference point\((t=0)\) and units of time. The number of cells in a tumor doubles every 6 weeks starting with 8 cells. After how many weeks does the tumor have 1500 cells?

Short answer

Answer: Approximately 25.6 weeks.

Step-by-step solution

Identify the initial number of cells (a)

Since the tumor starts with 8 cells, the initial number of cells (a) is 8.

Determine the growth factor (b)

Since the number of cells doubles every 6 weeks, the growth factor (b) is 2.

Calculate the growth rate (k)

As the number of cells doubles every 6 weeks, we can set up the exponential growth function with time t=6 weeks to find the growth rate (k): 8 * 2^(6k) = 16 (Since the number of cells doubles) Now, divide both sides of the equation by 8: 2^(6k) = 2 Taking the logarithm of both sides: 6k = log_2(2) 6k = 1 Now, divide both sides by 6: k = 1/6

Write the exponential growth function

Now that we have the values for a, b, and k, we can write the exponential growth function: y(t) = 8 * 2^(t/6)

Find the time when the number of cells is 1500

To find the time when the number of cells is 1500, we plug in y(t) = 1500 into the exponential growth function: 1500 = 8 * 2^(t/6) Now, divide both sides by 8: 187.5 = 2^(t/6) Take the logarithm of both sides: log_2(187.5) = t/6 Now, multiply both sides by 6 to find t: t = 6 * log_2(187.5) Calculate the value for t: t ≈ 25.6 weeks The tumor has 1500 cells after approximately 25.6 weeks.

Key concepts

Growth Factor

In exponential growth, the growth factor is a vital element. It determines how much a quantity increases over a specific time frame. For this problem, the growth factor is 2 because the tumor doubles in size every 6 weeks.
This means that the number of cells in the tumor is multiplying by a factor of 2 every 6 weeks:

• A growth factor greater than 1 signifies growth.
• A growth factor less than 1 would indicate decay.

This concept is essential to understand exponential functions and how quantities change rapidly over time. The growth factor helps us calculate future sizes or counts using the exponential growth formula.

Growth Rate

The growth rate is a measure that quantifies the speed of growth during each period. To find the growth rate in this exercise, we first recognize the doubling pattern of the tumor's growth.
The growth rate is derived from the relationship between growth factor and time period. Using the equation through logarithms, we set up:

• \(2^{6k} = 2 \)
• The logarithm helps solve for \(k\), which is the growth rate.
• In this instance, \(k = \frac{1}{6}\) per week.

The growth rate is an incremental value indicating weekly changes, bridging our understanding of how fast or slow growth happens.

Exponential Function

An exponential function models situations where growth (or decay) happens rapidly. It takes the form \(y(t) = a \cdot b^{kt}\), where:

• \(y(t)\) is the quantity at time \(t\).
• \(a\) is the initial quantity.
• \(b\) is the growth factor.
• \(k\) is the growth rate.

This problem involves finding the number of cells in a tumor over time. Using the initial value, growth factor, and growth rate, we defined the function:
\(y(t) = 8 \cdot 2^{t/6}\)
This function allows calculation of how many cells exist at any given time \(t\), provided units are consistent.

Logarithms

Logarithms are a mathematical tool that helps solve equations where the unknown variable is an exponent, such as in exponential functions. By converting the multiplication of exponents into an addition problem, logarithms simplify solving exponential growth problems.
In this exercise, we used logarithms to find the time required for the tumor to grow to 1500 cells:

• Start with the equation: \(2^{t/6} = 187.5\)
• Apply logarithms: \(\log_2(187.5) = \frac{t}{6}\)
• This technique allows us to isolate \(t\), ensuring we calculate it accurately.

Logarithms thus empower us to handle complex growth problems by transforming them into more manageable forms.