Solve for $$ x $$ $$ \ln 2+\ln (x+3)=-3 $$

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Solve the equation $$ \ln(2)+\ln(x+3)=-3 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\ln{\left(2ight)}+\ln{\left(x+3ight)}=-3$$
- step1: Find the domain:
  $$\ln{\left(2ight)}+\ln{\left(x+3ight)}=-3,x>-3$$
- step2: Transform the expression:
  $$\ln{\left(2\left(x+3ight)ight)}=-3$$
- step3: Convert the logarithm into exponential form:
  $$2\left(x+3ight)=e^{-3}$$
- step4: Evaluate the power:
  $$2\left(x+3ight)=\frac{1}{e^{3}}$$
- step5: Multiply by the reciprocal:
  $$2\left(x+3ight)	imes \frac{1}{2}=\frac{1}{e^{3}}	imes \frac{1}{2}$$
- step6: Multiply:
  $$x+3=\frac{1}{2e^{3}}$$
- step7: Move the constant to the right side:
  $$x=\frac{1}{2e^{3}}-3$$
- step8: Subtract the numbers:
  $$x=\frac{1-6e^{3}}{2e^{3}}$$
- step9: Check if the solution is in the defined range:
  $$x=\frac{1-6e^{3}}{2e^{3}},x>-3$$
- step10: Find the intersection:
  $$x=\frac{1-6e^{3}}{2e^{3}}$$
The solution to the equation $$ \ln 2 + \ln (x+3) = -3 $$ is $$ x = \frac{1-6e^{3}}{2e^{3}} $$.
$$ x = \frac{1 - 6e^{3}}{2e^{3}} $$

Solve for \( x \) \[ \ln 2+\ln (x+3)=-3 \]

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