Question 1128878
{{{ P = P[o]e^(kt) }}}<br>
{{{ 2P[o] = P[o]e^(0.012t) }}}<br>
Cancelling the {{{P[o]}}}'s and taking log of both sides:
{{{ ln(2) = 0.012t }}}<br>
{{{ t = ln(2)/0.012 = 0.693/0.012 = highlight( 57.8 ) }}} years