5=12.4te-0.2t.
Let f(t)=12.4te-0.2t-5.
f'(t)=12.4e-0.2t+12.4t(-0.2e-0.2t)=12.4e-0.2t(1-0.2t).
Iterative method (Newton):
tn+1=tn-f(tn)/f'(tn), where t0=1.
t1=0.4032..., t2=0.44006..., t3=0.4403..., t4=0.44034848066, t5=t4.
Therefore t converges to t=0.44035 approx.