How many years (to two decimal places) will it take an investment of $35,000 to grow to $50,000 if it is invested at 4.75% compounded continuously?
The formula for interest compounded continuously is:
`FV = Pe^(rt)`
where FV - future value P - principal amount
r - interest rate per year t - number of years
To solve for t, substiute values of FV=$50000, P=$35000 and r=4.75% (in decimal form, 0.0475).
`50000 = 35000 e^(0.0475t)`
`50000/35000 = e^(0.0475t)`
Reduce the fraction at the left side to its lowest term.
`10/7 = e^(0.0475t)`
Take the natural logarithm of both sides.
`ln (10/7) = ln e ^(0.0475t)`
Apply the power rule for logarithm which is `log_b a^n = n log_ba` .
`ln (10/7) = (0.0475t) lne`
Also in logarithm, if the base and the argument are the same, the resulting value is 1 ( `log_b b = 1` ). Note that ln e ` `is the same as `log_e e` .
`ln(10/7) = (0.0475t)*1`
`ln (10/7) = 0.0475t`
`ln(10/7)/0.0475 = t`
` 7.51 = t `
At an interest rate of 4.75% compunded continuously, it takes 7.51 years for an investment to increase from $35000 to $50000.