Question 1185576
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How long would it take $1000, invested at 6% annually compounded continuously, to double in value?
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<pre>
    2000 = {{{1000*e^(0.06*t)}}}


    {{{2000/1000}}} = {{{e^(0.06*t)}}}


    2 = {{{e^(0.06*t)}}}


    ln(2) = 0.06*t


    t = {{{ln(2)/0.06}}} = 11.552 years


<U>ANSWER</U>.  11.552year,  or  11 years and 202 days;  11 years and about 7 months.
</pre>

Solved.


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To see many other similar solved problems on continuously compounded accounts, &nbsp;see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/logarithm/Problems-on-continuously-compound-accounts.lesson>Problems on continuously compounded accounts</A> 

in this site.


After reading this lesson, &nbsp;you will tackle such problems on your own without asking for help from outside.



Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic "<U>Logarithms</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.