find the least number of complete years in which a sum of money put out at 45 % compound interest will be more than double of itself?  Choose the correct choice from the following answers:
A) 2 B) 3 C) 4 D) 5 E) 6
<gadget id="calculator">log(2)</gadget>
<output>log(2) = around 0.693147</output>

<gadget id="calculator">45 / 100</gadget>
<output>9/20 = around 0.45</output>

<gadget id="calculator">1 + (9/20)</gadget>
<output>29/20 = around 1.45</output>

<gadget id="calculator">log(29/20)</gadget>
<output>log(29/20) = around 0.371564</output>

<gadget id="calculator">log(2) / log(29/20)</gadget>
<output>log(2)/log(29/20) = around 1.865488</output>

<gadget id="calculator">(log(2)/log(29/20)) + 1</gadget>
<output>1 + log(2)/log(29/20) = around 2.865488</output>

<gadget id="calculator">floor(1 + log(2)/log(29/20))</gadget>
<output>2</output>

<result>A</result>
find the least number of complete years in which a sum of money put out at 45 % compound interest will be more than double of itself?
floor(add(divide(log(const_2), log(add(const_1, divide(45, const_100)))), const_1))
"2 years answer : a"