Mod Signs

[abeybaby]

the easiest way is to solve for t and then input your initial condition:
t = 2loge(|v-4|)+c
e^[(t-c)/2]=|v-4|
|v-4|=e^[-c/2] x e^(t/2)
v-4=+/- e^[-c/2] x e^(t/2). since e^[-c/2] is constant, we can say: Let +/-e^[-c/2] = A  (whether A is positive or negative will depend on whether v-4 is was positive or negative).

therefore, v=4+Ae^(t/2). v=0 when t=0. so:
0=4+A and A=-4.
so v=4-4e^(t/2) and if you went back to the log form, that would be: -(v-4)/4=e^(t/2) so 2loge(-(v-4)/4)=t and dropping the modulus would have been wrong in this case.