Math

Date: ___________ Page: ___________

The present value of an iPhone is [tex]$2250. If its value depreciates yearly by $[/tex]8\%[tex]$, what does $[/tex]V_0[tex]$ denote in the formula $[/tex]V_t = V_0 \left(1 - \frac{p}{100}\right)^t[tex]$ for calculating the value over time? Calculate the price of an iPhone after 4 years.

After how many years will the price of the iPhone be $[/tex]1752.048$?



Answer :

Certainly! Let's go through the problem step by step.

### Part 1: Understanding and Setting Up the Depreciation Formula

We have the following information:
- Present value ([tex]\( V_0 \)[/tex]) of the iPhone: [tex]\( 2250 \)[/tex] RS
- Annual depreciation rate ([tex]\( p \)[/tex]): [tex]\( 8\% \)[/tex]

The formula for the value of a depreciating asset over [tex]\( t \)[/tex] years is given by:
[tex]\[ NT = V_0 \times \left(1-\frac{p}{100}\right)^t \][/tex]

Here, [tex]\( NT \)[/tex] is the value of the iPhone after [tex]\( t \)[/tex] years.

### Part 2: Calculating the Price of the iPhone After 4 Years

We need to find the value of the iPhone after [tex]\( 4 \)[/tex] years. Using the given formula:

1. Present value [tex]\( V_0 = 2250 \)[/tex] RS
2. Depreciation rate [tex]\( p = 8\% \)[/tex]
3. Number of years [tex]\( t = 4 \)[/tex]

Let's calculate it step by step:

[tex]\[ NT = 2250 \times \left(1 - \frac{8}{100}\right)^4 \][/tex]

First, compute [tex]\( \left(1 - \frac{8}{100}\right) \)[/tex]:

[tex]\[ \left(1 - \frac{8}{100}\right) = 0.92 \][/tex]

Next, raise it to the power of [tex]\( 4 \)[/tex]:

[tex]\[ 0.92^4 \approx 0.7164 \][/tex]

Finally, multiply by the present value [tex]\( 2250 \)[/tex]:

[tex]\[ NT = 2250 \times 0.7164 \approx 1611.88 \][/tex]

So, the price of the iPhone after 4 years will be approximately [tex]\( 1611.88 \)[/tex] RS.

### Part 3: Finding the Number of Years to Reach a Target Value

Next, we need to find out after how many years the price of the iPhone will be [tex]\( 175204.8 \)[/tex] RS. We use the same formula but solve for [tex]\( t \)[/tex]:

[tex]\[ 175204.8 = 2250 \times \left(1 - \frac{8}{100}\right)^t \][/tex]

First, isolate the depreciation factor:

[tex]\[ \left(1 - \frac{8}{100}\right)^t = \frac{175204.8}{2250} \][/tex]

Compute the fraction:

[tex]\[ \frac{175204.8}{2250} \approx 77.8692 \][/tex]

Next, we need to solve for [tex]\( t \)[/tex]. We use the natural logarithm to do this:

[tex]\[ \left(0.92\right)^t = 77.8692 \][/tex]

Taking the natural logarithm of both sides gives:

[tex]\[ \ln\left(0.92^t\right) = \ln\left(77.8692\right) \][/tex]

Using the properties of logarithms:

[tex]\[ t \cdot \ln(0.92) = \ln(77.8692) \][/tex]

Now isolate [tex]\( t \)[/tex]:

[tex]\[ t = \frac{\ln(77.8692)}{\ln(0.92)} \][/tex]

Evaluating the logarithms:

[tex]\[ \ln(77.8692) \approx 4.3537 \][/tex]
[tex]\[ \ln(0.92) \approx -0.0838 \][/tex]

So:

[tex]\[ t = \frac{4.3537}{-0.0838} \approx -52.23 \][/tex]

This negative value indicates that it is not possible for the iPhone's price to depreciate to [tex]\( 175204.8 \)[/tex] RS, reflecting an error in initial assumptions or an exceptional external factor not considered by simple depreciation. In practice, such high future values due to depreciation calculations often mean appreciating value or external factors affecting pricing.

### Summary

1. The price of the iPhone after 4 years will be approximately [tex]\( 1611.88 \)[/tex] RS.
2. It will take approximately [tex]\(-52.23\)[/tex] years for the iPhone price to reach [tex]\( 175204.8 \)[/tex] RS, which is practically impossible given only depreciation. Additional factors or appreciation may be needed.