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.) ...?

Respuesta :

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 :

[tex]A=Pe^{Rt}[/tex]

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:

[tex]A=Pe^{Rt}[/tex]

[tex]\$ 2000=\$ 1000\times e^{0.075\times t}[/tex]

[tex]2=e^{0.075t}[/tex]

Taking logarithm on both side, we get:

[tex]\ln 2=\log e^{0.075t}[/tex]

[tex]\ln 2=0.075\times t[/tex]

[tex]t=9.24years[/tex]

Therefore, the time taken will be, 9.24 years