# Lesson 15

Using Graphs and Logarithms to Solve Problems (Part 1)

### Problem 1

The equation \(p(h) = 5,\!000 \boldcdot 2^h\) represents a bacteria population as a function of time in hours. Here is a graph of the function \(p\).

- Use the graph to determine when the population will reach 100,000.
- Explain why \(\log_{2} 20\) also tells us when the population will reach 100,000.

### Solution

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### Problem 2

*Technology required.* Population growth in the U.S. between 1800 and 1850, in millions, can be represented by the function \(f\), defined by \(f(t) = 5 \boldcdot e^{(0.028t)}\).

- What was the U.S. population in 1800?
- Use graphing technology to graph the equations \(y=f(t)\) and \(y=20\). Adjust the graphing window to the following boundaries: \(0< x< 100\) and \(0< y< 40\).
- What is the point of intersection of the two graphs, and what does it mean in this situation?

### Solution

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### Problem 3

The growth of a bacteria population is modeled by the equation \(p(h) = 1,\!000 e^{(0.4h)}\). For each question, explain or show how you know.

- How long does it take for the population to double?
- How long does it take for the population to reach 1,000,000?

### Solution

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### Problem 4

What value of \(b\) makes each equation true?

- \(\log_b 144 = 2\)
- \(\log_b 64 = 2\)
- \(\log_b 64 = 3\)
- \(\log_b 64 = 6\)
- \(\log_b \frac19 = \text-2\)

### Solution

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(From Unit 4, Lesson 10.)### Problem 5

Put the following expressions in order, from least to greatest.

- \(\log_2 11\)
- \(\log_3 5\)
- \(\log_5 25\)
- \(\log_{10} 1,\!000\)
- \(\log_2 5\)

### Solution

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(From Unit 4, Lesson 11.)### Problem 6

Solve \(9 \boldcdot 10^{(0.2t)} = 900\). Show your reasoning.

### Solution

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(From Unit 4, Lesson 14.)### Problem 7

Explain why \(\ln 4\) is greater than 1 but is less than 2.

### Solution

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(From Unit 4, Lesson 14.)