Question 1198606


{{{D=10e^(-0.9h) }}}

When the number of milligrams reaches {{{2}}}, means {{{D=2}}}

{{{2=10e^(-0.9h) }}}

{{{ln(2)=ln(10e^(-0.9h) )}}}

{{{ln(2)=ln(10)+ln(e^(-0.9h) )}}}

{{{ln(2)-ln(2*5)=ln(e^(-0.9h) )}}}

{{{ln(2)-(ln(2)+ln(5))=(-0.9h)ln(e)}}}

{{{-log(5)=-0.9h}}}

{{{h=-log(5)/-0.9}}}

{{{h=1.788264347149}}}

{{{h=1.8 }}}hours