I have no idea how to even begin writing this out. I need help desperately.
A student deposits $6,000 in a savings account with x% continuously compounded interest. How many years must he wait until the balance has doubled?
A = Pe^xt
where x is the rate and t is the # years.
When the amount has doubled, A=2P, so
2 = e^xt
ln2 = xt
t = ln2/x
To solve this problem, you need to use the formula for compound interest:
A = P * e^(rt)
Where:
A = final amount or balance
P = principal amount (initial deposit)
e = base of the natural logarithm (approximately equal to 2.71828)
r = interest rate (in decimal form)
t = time (in years)
In this case, the value of A is double the initial deposit (2 * $6,000 = $12,000). We need to find out how many years (t) it will take for the balance to double.
The equation becomes:
$12,000 = $6,000 * e^(r*t)
Now, we can solve for t.
To do this, we need to rearrange the equation:
e^(r*t) = 2
Take the natural logarithm (ln) of both sides to isolate the exponent:
ln(e^(r*t)) = ln(2)
Using the property of logarithms, the exponent comes down as a coefficient:
r*t * ln(e) = ln(2)
Since ln(e) is equal to 1, the equation simplifies to:
r*t = ln(2)
Finally, solve for t by dividing both sides by r:
t = ln(2) / r
Now, you know that the time required to double the balance is equal to the natural logarithm of 2 divided by the interest rate.