How long would it take to earn $7000 on a 3500 dollar investment at rate of 7.8% compounded continuously?
Would it be something like this:
7000=3000*(1+0.078)/? not sure about what to divided by???
Help-I'm really confused!!!!
dx/dt =.078 x
dx/x = .078 dt
ln x = .078 t
x =X e^(.078 t)
7000 = 3500 e^(.078 t)
ln 2 = .078 t
t = .693/.078 = 8.9 years
To calculate the time it takes to earn $7000 on a $3500 investment at a continuously compounded interest rate of 7.8%, you can use the formula for compound interest:
A = P * e^(rt),
where A is the final amount, P is the initial principal, e is the base of the natural logarithm (approximately 2.71828), r is the interest rate, and t is the time in years.
In this case, you have:
A = $7000,
P = $3500,
r = 7.8%.
First, convert the interest rate to decimal form by dividing it by 100:
r = 7.8% / 100 = 0.078.
Now, substitute the values into the formula:
$7000 = $3500 * e^(0.078t).
Next, isolate the exponential term:
e^(0.078t) = $7000 / $3500 = 2.
To solve for t, you need to take the natural logarithm (ln) of both sides to remove the exponential term. The natural logarithm is the inverse function of the base e.
ln(e^(0.078t)) = ln(2).
Applying the properties of logarithms, the power can be brought down as a coefficient:
0.078t * ln(e) = ln(2).
Since ln(e) is equal to 1, the equation simplifies to:
0.078t = ln(2).
Finally, solve for t by dividing both sides by 0.078:
t = ln(2) / 0.078.
Using a calculator, find the value of ln(2) (approximately 0.6931), and divide it by 0.078:
t ≈ 0.6931 / 0.078 ≈ 8.88.
Therefore, it would take approximately 8.88 years to earn $7000 on a $3500 investment with a continuously compounded interest rate of 7.8%.