Answer:
6 years
Step-by-step explanation:
Given
[tex]Initial\ Investment = 150[/tex]
[tex]Ratio = double[/tex]
Required
When will it get to 4000
This question illustrates a geometric progression and will be solved using:
[tex]T_n = ar^{n-1}[/tex]
In this case;
[tex]T_n = 4000[/tex]
[tex]r = double = 2[/tex]
[tex]a = 150[/tex]
So, we're to solve for n
[tex]4000 = 150 * 2^{n - 1}[/tex]
Divide through by 150
[tex]\frac{4000}{150} = 2^{n-1}[/tex]
[tex]26.67= 2^{n-1}[/tex]
Take Log of both sides
[tex]Log26.67= Log2^{n-1}[/tex]
Apply law of logarithm
[tex]Log26.67= ({n-1})Log2[/tex]
Take Log values
[tex]1.42602301569 = (n - 1) * 0.30102999566[/tex]
Make n - 1 the subject
[tex]n - 1 = \frac{1.42602301569}{0.30102999566}[/tex]
[tex]n - 1 = 4.73714591984[/tex]
Add 2 to both sides
[tex]n = 1 + 4.73714591984[/tex]
[tex]n = 5.73714591984[/tex]
[tex]n = 6[/tex] (Approximated)
Hence, it'll take 6 years
