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**Growth or Decay Functions:**

1. Determine whether each function shows growth or decay, then graph:

- [tex] f(x) = 0.4^x [/tex]
- [tex] f(x) = 1.3 \left(\frac{2}{5}\right)^x [/tex]
- [tex] f(x) = \frac{7}{8}(1.1)^x [/tex]
- [tex] f(x) = 50(1+0.04)^x [/tex]

2. Gina buys a car for [tex] \$13,500 [/tex]. Assume that its value will decrease by about [tex] 15\% [/tex] per year. Write an exponential function to model the value of the car and graph the function. When will the value fall below [tex] \$3,000 [/tex]?

**Graphing Functions and Their Inverses:**

3. Graph each function. Then write and graph its inverse:

- [tex] f(x) = x - 1.06 [/tex]
- [tex] f(x) = \frac{5}{6}x - 1.06 [/tex]
- [tex] f(x) = 1.06 - \frac{5}{6}x [/tex]
- [tex] f(x) = \frac{1}{4}\left(1.06 - \frac{5}{6}x\right) [/tex]

**Alternative Form (Exponential or Logarithmic):**

4. Write each expression in the alternative form:

- [tex] 16^{\frac{1}{4}} = 2 [/tex]
- [tex] 16^{-0.5} = \frac{1}{4} [/tex]
- [tex] \log_{\frac{1}{6}} 64 = -3 [/tex]
- [tex] \log_{81} \frac{1}{3} = -\frac{1}{4} [/tex]

**Function Inverses:**

5. Use the given [tex] x [/tex]-values to graph each function. Then write and graph its inverse. Describe the domain and range of the inverse function:

- [tex] f(x) = \left(\frac{1}{4}\right)^x ; x = -1, 0, 2, 4 [/tex]
- [tex] f(x) = 2.5^x ; x = -1, 0, 1, 2, 3 [/tex]
- [tex] f(x) = 5 - 3 ; x = -1, 0, 1, 2 [/tex]

**Simplification:**

6. Simplify the following logarithmic expressions:

- [tex] \log_4 128 - \log_4 8 [/tex]
- [tex] \log_2 12.8 + \log_2 5 [/tex]
- [tex] \log_3 243^2 [/tex]
- [tex] 5^{\log_3 x} [/tex]

**Solving Equations:**

7. Solve the following equations:

- [tex] 3^{x-x} = 729^{\frac{x}{2}} [/tex]
- [tex] 5^{1.5-x} \leq 25 [/tex]
- [tex] \log_4(x+48) = 3 [/tex]
- [tex] \log(6x^2) - \log 2x = 1 [/tex]

**Application Problems:**

8. The rate at which a liquid vitamin breaks down in the average human body can be modeled by [tex] y = D(0.95)^x [/tex], where [tex] y [/tex] ml of the original dose [tex] D [/tex] remains after [tex] x [/tex] minutes. How long will it take for an original dose of 15 ml to be reduced to less than 5 ml?

9. Plutonium Pu-239 has a half-life of about 24,000 years. The formula [tex] \frac{1}{2}=e^{-kt} [/tex] relates the half-life [tex] t [/tex] to the decay constant [tex] k [/tex] for a given substance. How much of a 100-gram quantity of plutonium will remain after 5 years?

10. The function [tex] f(x) = \ln x [/tex] is shifted 2 units left and 1 unit up and is vertically stretched by a factor of 3. Write the transformed function.

11. Use logarithmic regression to find the function that models the population data in the table. In what year will the population exceed 100?

\[
\begin{tabular}{|c|c|c|c|}
\hline
Population & 50 & 62 & 78 \\
\hline
Year & 1 & 2 & 3 \\
\hline
\end{tabular}
\]

Answer :

To solve the problem of finding out how long it will take for a liquid vitamin to be reduced from an original dose of 15 ml to less than 5 ml, where the rate of breakdown is modeled by the formula [tex]\( y = D(0.95)^x \)[/tex], follow these steps:

1. Set Up the Equation:
We start by substituting the given values into the equation:
[tex]\[
5 = 15 \times (0.95)^x
\][/tex]

2. Solve for [tex]\((0.95)^x\)[/tex]:
Divide both sides by 15 to isolate [tex]\((0.95)^x\)[/tex]:
[tex]\[
(0.95)^x = \frac{5}{15} = \frac{1}{3}
\][/tex]

3. Use Logarithms to Solve for [tex]\(x\)[/tex]:
Take the natural logarithm (ln) of both sides to make use of the logarithmic property that allows us to bring the exponent down:
[tex]\[
\ln((0.95)^x) = \ln\left(\frac{1}{3}\right)
\][/tex]
[tex]\[
x \cdot \ln(0.95) = \ln\left(\frac{1}{3}\right)
\][/tex]

4. Isolate [tex]\(x\)[/tex]:
Solve for [tex]\(x\)[/tex] by dividing both sides by [tex]\(\ln(0.95)\)[/tex]:
[tex]\[
x = \frac{\ln\left(\frac{1}{3}\right)}{\ln(0.95)}
\][/tex]

Upon evaluating this expression (using a calculator):

- [tex]\(\ln\left(\frac{1}{3}\right) \approx -1.0986\)[/tex]
- [tex]\(\ln(0.95) \approx -0.0513\)[/tex]

We find:
[tex]\[
x \approx \frac{-1.0986}{-0.0513} \approx 21.42
\][/tex]

So, it will take approximately 21.42 minutes for the amount of vitamin to be reduced to less than 5 ml.

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