Question

what amount of time in months is necessary for a principal of $6,000 to produce $550 and interest at 10%?

Answer 1

the compound interest formula is given by

`A=P(1+(r)/(n))^(nt)`

where A is the amount you will have, P is the principal, r is the annual interest rate, n is the amount of times the interest is compounded per time period, and t is the amount of time.

In our case, we need to find the time t. Then, by moving the Principal to the left hand side, we get

`(A)/(P)=(1+(r)/(n))^(nt)`

By applying natural logarithm on both sides, we get

`\ln ((A)/(P))=nt\cdot ln(1+(r)/(n))`

now, we can isolate t as

`t=(\ln ((A)/(P)))/(n\ln (1+(r)/(n)))`

Now, we can substitute our given values into this expression. It yields,

`t=(\ln ((6550)/(6000)))/(12\ln (1+(0.1)/(12)))`

which gives

`t=(0.0877)/(12(0.0083))`

then, the time (in years) is

`t=0.88\text{ years}`

Now, we must convert this result in months. Since 1 year has 12 months, we have

`\begin{gathered} t=0.88*12 \\ t=10.56\text{ months} \end{gathered}`

that is, the answer is 10.56 months