1 e^.012t = 2
take ln of both sides
ln e^.01t = ln2
.012t ln e = ln 2 , but ln e = 1
.12t = ln 2
t = ln2/.012 = 57.76 years
or 58 years to the nearest year.
I got 6 years, is this correct?
take ln of both sides
ln e^.01t = ln2
.012t ln e = ln 2 , but ln e = 1
.12t = ln 2
t = ln2/.012 = 57.76 years
or 58 years to the nearest year.
A = P * e^(rt)
Where:
A is the final amount (double the initial investment)
P is the principal (initial investment)
e is the mathematical constant approximately equal to 2.71828
r is the interest rate, expressed as a decimal (1.2% = 0.012)
t is the time in years we want to find
In this case, we want to find t, the number of years it will take for the investment to double. So we can rewrite the equation as:
2P = P * e^(0.012t)
Dividing both sides by P, we get:
2 = e^(0.012t)
To isolate t, we take the natural logarithm (ln) of both sides:
ln(2) = ln(e^(0.012t))
Since ln(e^x) = x, we can simplify further:
ln(2) = 0.012t
Now, we can solve for t by dividing both sides by 0.012:
t = ln(2) / 0.012
Using a calculator, we find that ln(2) ≈ 0.693. Hence, the calculation becomes:
t = 0.693 / 0.012
Simplifying, we get:
t ≈ 57.75
Rounding this value to the nearest thousandth gives approximately 57.750.
Therefore, it will take approximately 57.750 years for your investment to double in value in a continuously compounding account, rounded to the nearest year.
Hence, your initial estimate of 6 years is not correct.