Question 362272
The amount at time {{{t=0}}} is {{{Q(0)=1800}}}
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Find {{{t}}} when {{{Q(t)=(1/2)(1800)=900}}}
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{{{900=1800e^(-0.025t)}}}
{{{e^(-0.025t)=0.5}}}
{{{-0.025t=ln(0.5)}}}
{{{t=-ln(0.5)/(0.025)}}} or approximately,
{{{t=27.73}}}