Question 1161309
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<pre>

The basic (starting) equation is

    2 = {{{(1+0.07/4)^n}}}

where n is the number of compound periods (quarters).



Take the logarithm base 10 of both sides

    log(2) = {{{n*log((1+0.07/4))}}}

    n = {{{log((2))/log((1+0.07/4))}}} = 39.95 = 40 quartes (rounded).


40 quarters = 10 years.


<U>ANSWER</U>.  10 years.
</pre>

Solved.


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To see many other similar solved problems, look into the lesson

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