Question 1153872
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<pre>

1500 = {{{1000*e^(0.05*t)}}},


t is the time in years;  "e" is the base of natural logarithm,  e = 2.71828... (approx.)


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{{{1500/1000}}} = {{{e^(0.05*t)}}}


1.5 = {{{e^(0.05*t)}}}


Take the natural logarithm of both sides


ln(1.5) = 0.05*t


t = {{{ln(1.5)/0.05}}} = 8.109 = 8.11 years (rounded as required).    <U>ANSWER</U>
</pre>

SOLVED.


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It is a standard way to solve such problems (with continuous compounding).


MEMORIZE it as a MANTRA.



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 compound accounts</A> 

in this site.


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 lessons are 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.