Question 506930


{{{A=Pe^(rt)}}} Start with the continuous compounding formula.



{{{1000000=1000*e^(0.09*t)}}} Plug in {{{A=1000000}}}, {{{P=1000}}}, and {{{r=0.09}}} (the decimal equivalent of 9%).



{{{1000000/1000=e^(0.09*t)}}} Divide both sides by {{{1000}}}.



{{{1000=e^(0.09*t)}}} Evaluate {{{1000000/1000}}} to get {{{1000}}}.



{{{ln(1000)=ln(e^(0.09*t))}}} Take the natural log of both sides.



{{{ln(1000)=0.09*t*ln(e)}}} Pull down the exponent using the identity {{{ln(x^y)=y*ln(x))}}}.



{{{ln(1000)=0.09*t*1}}} Evaluate the natural log of 'e' to get 1.



{{{ln(1000)=0.09*t}}} Multiply and simplify.



{{{6.90775527898214=0.09*t}}} Evaluate the natural log of {{{1000}}} to get {{{6.90775527898214}}} (this value is approximate).



{{{6.90775527898214/0.09=t}}} Divide both sides by {{{0.09}}} to isolate 't'.



{{{76.7528364331349=t}}} Evaluate {{{6.90775527898214/0.09}}} to get {{{76.7528364331349}}}.



{{{t=76.7528364331349}}} Flip the equation.



{{{t=77}}} Round to the nearest whole year.



So it will take roughly 77 years.


Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>
or you can visit my website here: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a>


Thanks,


Jim