How long will it take for an investment of $3000 to double in value if the interest rate is 7.5% per year, compounded continuously? (Round your answer to two decimal places.) ...?
A = $6000 P =$3000, r = 7.5% , t = ? years
t = log(A/P) / r
t = 9.24 years
Answer : The time taken will be, 9.24 years
Explanation : Given,
Principle = $1000
Rate = 7.5 %
Formula used :
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where,
A = amount = 2P = $2000
P = principle = $1000
R = interest rate = 7.5 % = 0.075
T = time = ?
Now put all the given values in the above formula, we get:
\(A=Pe^{Rt}\)
\(\$ 2000=\$ 1000\times e^{0.075\times t}\)
\(2=e^{0.075t}\)
Taking logarithm on both side, we get:
\(\ln 2=\log e^{0.075t}\)
\(\ln 2=0.075\times t\)
\(t=9.24years\)
Therefore, the time taken will be, 9.24 years