Question 345408
{{{2P=Pe^(0.21t)}}}
{{{e^(0.21t)=2}}}
{{{0.21t=ln(2)}}}
{{{t=ln(2)/0.21}}} or approximately,
{{{t=3.3}}} years
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{{{0.33t=ln(2)}}}
{{{t=ln(2)/0.33}}}
{{{t=2.1}}} years