Answer:
119.95 years
Step-by-step explanation:
The general equation is given by:
[tex]P = A*(1 + \frac{r}{n} )^{n*t}[/tex]
Where:
A is the initial amount, we know that the first deposit is of $150, then:
A = $150
t is the variable, in this case, is the number of years.
n = number of times that the interest is compounded in one unit of t, because the interest is compounded monthly, we have n = 12.
r = interest rate in decimal form.
r = 2.5%/100% = 0.025
Replacing these in our equation, we get that:
[tex]P = 150*(1 + \frac{0.025}{12} )^{12*t}[/tex]
Now we want to find the time such that his savings, P, are equal to $3000.
Then we need to solve the equation:
[tex]P = 150*(1 + \frac{0.025}{12} )^{12*t} = 3000[/tex]
[tex](1 + \frac{0.025}{12} )^{12*t} = 3000/150 = 20\\[/tex]
Now, remember that:
Ln(a^x) = x*ln(a)
So if we apply the natural logarithm to bot sides, we get:
[tex]Ln((1 + \frac{0.025}{12} )^{12*t}) = Ln( 20)\\\\(12*t)*Ln(1 + \frac{0.025}{12}) = Ln(20)\\\\t = \frac{Ln(20)}{12*Ln(1 + \frac{0.025}{12})} = 119.95[/tex]
So after 119.95 years you will have the $3000.
