$9000 is invested at 8% compounded quarterly. In how many years will the account have 20)
grown to $11,000? Round your answer to the nearest tenth of a year.
A = P(1 + r/n)^(nt)
11,000 = 9000(1 + 0.08/4)^(4t)
11,000 = 9000(1.02)^(4t)
11/9 = (1.02)^(4t)
Time to use logarithms.
log_1.02(11/9) = 4t
t = 2.5 years
To find the number of years it takes for an investment to grow from $9,000 to $11,000 at 8% compounded quarterly, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount (in this case, $11,000)
P = the initial principal (in this case, $9,000)
r = the interest rate per year (8% = 0.08)
n = the number of compounding periods per year (quarterly = 4)
t = the number of years
Plugging in the values, we have:
$11,000 = $9,000(1 + 0.08/4)^(4t)
Now, let's solve for t. Divide both sides of the equation by $9,000:
(11,000 / 9,000) = (1 + 0.08/4)^(4t)
Simplify the equation:
(11/9) = (1 + 0.02)^(4t)
To solve for t, take the natural logarithm of both sides of the equation:
ln(11/9) = ln(1.02)^(4t)
Using logarithm properties, we can bring the power of 4t down:
ln(11/9) = 4t * ln(1.02)
Now, divide both sides of the equation by 4 * ln(1.02):
t = ln(11/9) / (4 * ln(1.02))
Using a calculator, you can compute this expression and find the value of t.(rounded to the nearest tenth of a year)