Question 1154570
Given:
{{{ t = 1 }}} hr
{{{ .75A[0] = A[0]*e^(k*1 ) }}}
{{{ .75 = e^k }}}
{{{ k = ln( .75 ) }}}
{{{ k = -.28768 }}}
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{{{ .5A[0] = A[0]*e^( k*t ) }}}
{{{ .5 = e^( k*t ) }}}
{{{ ln( .5 ) = k*t }}}
{{{- .69315 =- .28768t }}}
{{{ t = 2.4094 }}}
{{{ t = 2.41 }}} ( rounded off )
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check the math & get a 2nd opinion if needed