Question

1. a patient is diagnosed with 30 malignant cancer cells. the cells are growing at a rate of 3% per day. find an exponential prediction model that fits this data and use the model to find the missing values of the chart. graph the chart values, sketch the graph.

$y = acdot b^{x}$

Explanation:

Step1: Determine the exponential model

The general form of an exponential model is \(y = A\cdot B^{x}\). When \(x = 0\), \(y=30\). Substituting into the formula \(y = A\cdot B^{x}\), we get \(30=A\cdot B^{0}\). Since \(B^{0}=1\) for \(B
eq0\), then \(A = 30\).

The growth rate is \(r=3\%=0.03\). For an exponential growth model \(y = A(1 + r)^{x}\), so \(B=1 + r=1.03\). The model is \(y = 30\times(1.03)^{x}\)

Step2: Find the value when \(x = 14\)

Substitute \(x = 14\) into the model \(y=30\times(1.03)^{14}\)

$$
LATEXBLOCK0
$$

Step3: Find the value when \(x = 40\)

Substitute \(x = 40\) into the model \(y=30\times(1.03)^{40}\)

$$
LATEXBLOCK1
$$

Step4: Find \(x\) when \(y = 300\)

Set \(y = 300\) in the model \(300=30\times(1.03)^{x}\)
First, divide both sides by \(30\): \(\frac{300}{30}=(1.03)^{x}\), so \(10=(1.03)^{x}\)
Take the natural logarithm of both sides: \(\ln(10)=\ln((1.03)^{x})\)
Using the property \(\ln(a^{b})=b\ln(a)\), we have \(\ln(10)=x\ln(1.03)\)

$$ x=\frac{\ln(10)}{\ln(1.03)}\approx\frac{2.302585}{0.0295598}\approx78 $$

Answer:

When \(x = 14\), \(y\approx45.38\); when \(x = 40\), \(y\approx97.86\); when \(y = 300\), \(x\approx78\)