Question

Peter purchases a baseball card for $150 as an investment. His research tells him that the value of

the card should double every 3 years. When will the card reach a value of $ 4000?

Answer 1

6 years

Explanation:

Given

`Initial\ Investment = 150`

`Ratio = double`

Required

When will it get to 4000

This question illustrates a geometric progression and will be solved using:

`T_n = ar^(n-1)`

In this case;

`T_n = 4000`

`r = double = 2`

`a = 150`

So, we're to solve for n

`4000 = 150 * 2^(n - 1)`

Divide through by 150

`(4000)/(150) = 2^(n-1)`

`26.67= 2^(n-1)`

Take Log of both sides

`Log26.67= Log2^(n-1)`

Apply law of logarithm

`Log26.67= ({n-1})Log2`

Take Log values

`1.42602301569 = (n - 1) * 0.30102999566`

Make n - 1 the subject

`n - 1 = (1.42602301569)/(0.30102999566)`

`n - 1 = 4.73714591984`

Add 2 to both sides

`n = 1 + 4.73714591984`

`n = 5.73714591984`

`n = 6` (Approximated)

Hence, it'll take 6 years