supoose 1000 is invested at a rate of 5% compounded quarterly, how long will it take the amount to double ?
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?
i dont have any idea of have to solve for x, that is where im stuck on
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 ?
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?
@dmparquet Do you have a calculator to work out the answer?
yes i do have im trying to solve
\[\ln 2\div \ln 1.0125=\]
the answer would be 55.74 ?
hold on i did bad
5.08?
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.
Number of years to double the $1000 = 56/4 = ?
im not understanding now
i just typed on my calculator the divison of ln2 and ln1.0125 ?
i just did what you said the answer is 14 ?
Then you should try another calculator for the log calculation. My calculation checks back correctly. Yes, the correct final answer is 14 years.
thanks man, i really apreciate it
can you help to solve this one real quick ? is my last one ? suppose a 1000$ 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
\[2000=1000\times e ^{5r}\] Taking logs of both sides: \[\ln 2000=\ln 1000+5r\] \[5r=\ln 2000-\ln 1000\] \[r=\frac{\ln 2000-\ln 1000}{5}\]
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