So I think I have got these down I just really want to double check the answers to these two word questions. Thanks
How long will it take for $9500 to grow to $36,300 at an interest rate of 12.4% if the interest is compounded continuously? Round the number of years to the nearest hundredth.
How long must $4000 be in a bank at 6% compounded annually to become $6757.92? (Round to the nearest year.)
What did you get?
10.81 yrs for the first question
9 yrs for the second
Both answers are correct. Good work!
To determine the answers to these questions, we can use the formula for compound interest:
A = P * e^(rt)
Where:
A is the final amount after interest (future value)
P is the principal amount (initial value)
e is the base of the natural logarithm (approximately 2.71828)
r is the interest rate (expressed as a decimal)
t is the time in years
For the first question:
P = $9500
A = $36,300
r = 12.4% = 0.124
We need to find t.
Plugging in the given values into the formula, we have:
$36,300 = $9500 * e^(0.124t)
To solve for t, divide both sides by $9500:
3.816 = e^(0.124t)
Take the natural logarithm of both sides to isolate t:
ln(3.816) = ln(e^(0.124t))
Using the logarithmic property ln(e^x) = x, we can simplify the equation:
ln(3.816) = 0.124t
Now, divide both sides by 0.124:
t = ln(3.816) / 0.124
Using a scientific or graphing calculator, evaluate the right side of the equation:
t ≈ 13.16 years
Rounding to the nearest hundredth, the answer is approximately 13.16 years.
For the second question:
P = $4000
A = $6757.92
r = 6% = 0.06
We again need to find t.
Plugging in the given values, we have:
$6757.92 = $4000 * (1+0.06)^t
Divide both sides by $4000:
1.68948 = (1.06)^t
To isolate t, take the logarithm of both sides, using the logarithmic property log(base a) a^x = x:
log(1.68948) = log((1.06)^t)
Now, divide both sides by log(1.06):
t = log(1.68948) / log(1.06)
Using a calculator, evaluate the right side of the equation:
t ≈ 6.78 years
Rounding to the nearest year, the answer is approximately 7 years.