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?

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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.**

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