7. ( 12 pts ) Solve a logarithmic equation. Don't forget to show your work. Solve for $$ x $$. Round your answer to four decimal places, if needed. $$ \ln \left(x^{2}+4 x-20\right)=0 $$

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Solve the equation $$ \ln(x^{2}+4x-20)=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\ln{\left(x^{2}+4x-20ight)}=0$$
- step1: Find the domain:
  $$\ln{\left(x^{2}+4x-20ight)}=0,x \in \left(-\infty,-2\sqrt{6}-2ight)\cup \left(2\sqrt{6}-2,+\inftyight)$$
- step2: Convert the logarithm into exponential form:
  $$x^{2}+4x-20=e^{0}$$
- step3: Evaluate the power:
  $$x^{2}+4x-20=1$$
- step4: Move the expression to the left side:
  $$x^{2}+4x-20-1=0$$
- step5: Subtract the numbers:
  $$x^{2}+4x-21=0$$
- step6: Factor the expression:
  $$\left(x-3ight)\left(x+7ight)=0$$
- step7: Separate into possible cases:
  $$\begin{align}&x-3=0\&x+7=0\end{align}$$
- step8: Solve the equation:
  $$\begin{align}&x=3\&x=-7\end{align}$$
- step9: Check if the solution is in the defined range:
  $$\begin{align}&x=3\&x=-7\end{align},x \in \left(-\infty,-2\sqrt{6}-2ight)\cup \left(2\sqrt{6}-2,+\inftyight)$$
- step10: Find the intersection:
  $$\begin{align}&x=3\&x=-7\end{align}$$
- step11: Rewrite:
  $$x_{1}=-7,x_{2}=3$$
The solution to the logarithmic equation $$ \ln \left(x^{2}+4x-20ight)=0 $$ is $$ x = -7 $$ or $$ x = 3 $$.

Therefore, the solutions for $$ x $$ are $$ x = -7 $$ and $$ x = 3 $$.
The solutions are $$ x = -7 $$ and $$ x = 3 $$.

( 12 pts ) Solve a logarithmic equation. Don’t forget to show your work.
Solve for . Round your answer to four decimal places, if needed.

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( 12 pts ) Solve a logarithmic equation. Don’t forget to show your work. Solve for . Round your answer to four decimal places, if needed.