Question 1175334
.
<pre>

A(t) = {{{P*(1 + r/2)^n}}},  where P is the principal  and  n is the number of compound periods.


1300 = {{{1000*(1+0.12/2)^n}}}


{{{1300/1000}}} = {{{1.06^n}}}


1.3 = {{{1.06^n}}}


log(1.3) = n*log(1.06)


n = {{{log((1.3))/log((1.06))}}} = 4.5  (approximately).


In order for the solution MAKES SENSE, it must be rounded to the NEAREST GREATER integer value
(to have INTEGER NUMBER of compounds).


THEREFORE, the correct  <U>ANSWER</U>  is 5 semi-annual compounding periods,

which is 2.5 years.
</pre>

Solved.


The answer by @ewatrrr is INCORRECT,

so do not accept it.


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To see many other similar solved problems, &nbsp;look into the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/percentage/lessons/Compound-interest-percentage-problem.lesson>Compounded interest percentage problems</A> 

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

in this site, and learn the subject from there.



After reading these lessons, 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 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


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