Simplifing logarithmic equation I have the result of a differential equation to be:
$$\ln(x+3)=3\ln(t+2)+C$$
I want this to be as simplified as possible. Can it be proceeded like:
$$e^{(x+3)}=3e^{(t+2)+C}$$
I am not sure about the equation above, but I have the final answer given to be:
$$x=-3+(t+2)^3C$$
Can somone please explain how this became the final answer?!
 A: You can imagine the constant as $\ln C'$, so you have:
$$\ln(x+3)=3\ln (t+2) + \ln C'$$
$$\ln(x+3)=\ln (t+2)^3 +\ln C'$$
$$\ln(x+3)=\ln C'(t+2)^3$$
Then, when you have natural logarithm on both sides of your equation, it means the arguments are equal (You'd write it as exponential terms otherwise):
$$x+3=C'(t+2)^3$$
Therefore:
$$x=C'(t+2)^3-3$$
A: Write $\ln(x+3)=3\ln(t+2)+C$ as $$ x+3 = \exp(\ln (t+2)^3 + C)=\exp(\ln(t+2)^3)\cdot \exp (C).$$ But this is the same as $$x + 3 = (t+2)^3\cdot K$$ where $K = \exp(C)$.
A: From $$\ln(x+3)=3\ln(t+2)+C$$
you cannot get $$e^{(x+3)}=e^{3(t+2)+C}$$ 
but instead $$e^{\ln(x+3)}=e^{3\ln(t+2)+C}$$
which can be written as $$x+3=(t+2)^3 \times e^C$$ and leads to the desired result: $e^C$ is a positive constant.
If you insisted, you could write $$e^{(x+3)}=e^{\left((t+2)^3 \times e^C\right)}$$ (note the power of $3$ rather than multiplying by $3$, and multiplying by the constant rather than adding) but it would not help.
A: let  our  C  some constant   $C=ln(c)$
now we have
  $ln(x+3)=3*lnc((t+2))$
so we have
  $x+3=c*(t+2)^3$ so  it means that
$x=c*(t+2)^3-3$
