Question

supoose 1000 is invested at a rate of 5% compounded quarterly, how long will it take the amount to double ?

Answer 1

The equation to solve for x is given below. Remember that x is the number of quarters for the $1000 to double.
\[2000=1000(1+\frac{0.05}{4})^{x}\]
Can you solve for x?

Answer 2

i dont have any idea of have to solve for x, that is where im stuck on

Answer 3

i do know how to find the logs but i just still dont know what to do with the x, obviously if i multiply log of 1012.5 * X will be log1012.5x ?

Answer 4

The equation can be simplified on the right hand side to give:
\[2000=1000\times (1.0125)^{x}\]
\[(1.0125)^{x}=2\]
\[x \times \ln 1.0125=\ln 2\]
\[x=\frac{\ln 2}{\ln 1.0125}\]
Can you find x, the number of quarters, now?

Answer 5

@dmparquet Do you have a calculator to work out the answer?

Answer 6

yes i do have im trying to solve

Answer 7

\[\ln 2\div \ln 1.0125=\]

Answer 8

the answer would be 55.74 ?

Answer 9

hold on i did bad

Answer 10

5.08?

Answer 11

Your first attempt was close. I get 55.79763 quarters. Round up to 56 quarters and divide 56 by 4 to find the number of years.

Answer 12

Number of years to double the $1000 = 56/4 = ?

Answer 13

im not understanding now

Answer 14

i just typed on my calculator the divison of ln2 and ln1.0125 ?

Answer 15

i just did what you said the answer is 14 ?

Answer 16

Then you should try another calculator for the log calculation. My calculation checks back correctly.
Yes, the correct final answer is 14 years.

Answer 17

thanks man, i really apreciate it

Answer 18

can you help to solve this one real quick ? is my last one ? suppose a 1000$ was invested in an account with interest compounded continously. If at the end of 5 years the account holds 2000$. what was the interest rate on the account ? i just dont know how to solve for r

Answer 19

\[2000=1000\times e ^{5r}\]
Taking logs of both sides:
\[\ln 2000=\ln 1000+5r\]
\[5r=\ln 2000-\ln 1000\]
\[r=\frac{\ln 2000-\ln 1000}{5}\]