.A car that sold for $ 28000 is depreciating at a rate of 7% per year. The value V(t) = 28000(0.93)^3. When will the car be worth half of its original selling price?                                 2. Use the graph of f(x)= In(x+2) to determine when f(x)=2                                          3. Solve for x: In(x+8) + In(3-x) = In10.                                          4. Solve for x: log_3x + 2log_3 4 =4.                                           5. Condensed form of log_3 12- log_3 2 + log_3 5

Let's address each of the questions one by one:

### 1. Car Depreciation Problem

The value of the car after $ t $ years is given by the function:
\[
V(t) = 28000(0.93)^t
\]
We need to determine when the car will be worth half of its original value, which is:
\[
\frac{28000}{2} = 14000
\]
So, we set up the equation:
\[
14000 = 28000(0.93)^t
\]
Dividing both sides by 28000:
\[
\frac{14000}{28000} = (0.93)^t
\]
\[
0.5 = (0.93)^t
\]
Taking the natural logarithm of both sides:
\[
\ln(0.5) = t \cdot \ln(0.93)
\]
Finally, solve for $ t $:
\[
t = \frac{\ln(0.5)}{\ln(0.93)} \approx 9.58 \text{ years}
\]
So, the car will be worth half of its original price after approximately 9.58 years.

### 2. Use the Graph of $ f(x) = \ln(x+2) $ to Determine When $ f(x) = 2 $

To solve $ f(x) = 2 $, we set up the equation:
\[
\ln(x + 2) = 2
\]
Exponentiating both sides to remove the logarithm:
\[
x + 2 = e^2
\]
\[
x = e^2 - 2
\]
So, $ x $ when $ f(x) = 2 $ is $ x = e^2 - 2 \approx 5.39 $.

### 3. Solve for $ x $: $ \ln(x+8) + \ln(3-x) = \ln 10 $

We start by combining the logarithms on the left side:
\[
\ln\left((x + 8)(3 - x)\right) = \ln 10
\]
Since the logarithms are equal, the arguments must also be equal:
\[
(x + 8)(3 - x) = 10
\]
Expanding the left side:
\[
3x - x^2 + 24 - 8x = 10
\]
Simplifying:
\[
-x^2 - 5x + 24 = 10
\]
\[
-x^2 - 5x + 14 = 0
\]
Multiply the equation by -1:
\[
x^2 + 5x - 14 = 0
\]
Now, solve this quadratic equation using the quadratic formula:
\[
x = \frac{-5 \pm \sqrt{5^2 + 4 \cdot 14}}{2} = \frac{-5 \pm \sqrt{25 + 56}}{2} = \frac{-5 \pm \sqrt{81}}{2}
\]
\[
x = \frac{-5 \pm 9}{2}
\]
The two possible solutions are:
\[
x = \frac{4}{2} = 2 \quad \text{or} \quad x = \frac{-14}{2} = -7
\]
We must check that these values are within the domain of the original logarithms. $ x = -7 $ is valid, but $ x = 2 $ leads to a negative argument inside $ \ln(3-x) $, which is not allowed. Therefore, the solution is:
\[
x = -7
\]

### 4. Solve for $ x $: $ \log_3 x + 2 \log_3 4 = 4 $

Use the logarithm properties to simplify:
\[
\log_3 x + \log_3 16 = 4
\]
\[
\log_3 (16x) = 4
\]
Now, exponentiate both sides with base 3:
\[
16x = 3^4
\]
\[
16x = 81
\]
Solve for $ x $:
\[
x = \frac{81}{16} = 5.0625
\]

### 5. Condensed Form of $ \log_3 12 - \log_3 2 + \log_3 5 $

Using logarithm properties:
\[
\log_3 \left(\frac{12}{2} \times 5\right) = \log_3 (6 \times 5) = \log_3 30
\]
So, the condensed form is:
\[
\log_3 30
\]

If you have any questions or need more details, feel free to ask!

### Follow-Up Questions:

1. Can you explain how to interpret the depreciation formula?
2. How does the value of $ e $ affect the solution in problem 2?
3. Why is the natural logarithm (ln) used in problem 3?
4. What is the significance of checking the domain in logarithmic equations?
5. Could you solve the quadratic equation in problem 3 using another method?
6. How do logarithmic properties simplify equations in problem 4?
7. What other properties of logarithms are useful in solving these types of problems?
8. How would you graphically represent the function in problem 2?

### Tip:

Always verify that the arguments of logarithmic functions remain positive, as logarithms are only defined for positive real numbers.

Explore solutions to complex mathematical problems including car depreciation and logarithmic equations. Learn about exponential decay and logarithmic functions with step-by-step explanations. Suitable for advanced high school students and anyone interested in applied mathematics.

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