Question 1032283
Substitute,
{{{u=T-38}}}
{{{du=dT}}}
So,
{{{du/dt=ku}}}
{{{du/u=kdt}}}
{{{ln(u)=kt+C}}}
{{{u=e^(kt+C)}}}
{{{u=Ce^(kt)}}}
{{{T-38=Ce^(kt)}}}
So when {{{t=0}}},
{{{T=75}}}
{{{T-38=C(1)}}}
{{{C=75-38}}}
{{{C=75}}}
and
when {{{t=30}}},
{{{T=60}}}
{{{60-38=75e^(30k)}}}
{{{e^(30k)=(60-38)/75}}}
{{{e^(30k)=22/75}}}
{{{30k=ln(22/75)}}}
{{{k=(1/30)ln(22/75)}}} 
So now the equation is complete.
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Find {{{T}}} when {{{t=60}}}
{{{T-38=75e^(60k)}}}
{{{T-38=75e^(2*ln(22/75))}}
{{{T-38=6.45)}}}
{{{highlight(T=44.5)}}}{{{F}}}
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Find {{{t}}} when {{{T=55}}}
{{{55-38=75e^(kt)}}}
{{{e^(kt)=17/75}}}
{{{kt=ln(17/75)}}}
{{{(1/30)ln(22/75)t=ln(17/75)}}}
{{{t=(30ln(17/75))/ln(22/75)}}}{{{min}}}
{{{highlight(t=36.3)}}}{{{min}}}