7. \( \ln 10+\ln x^{2}=10 \)

Solve the equation \( \ln(10)+\ln(x^{2})=10 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\ln{\left(10\right)}+\ln{\left(x^{2}\right)}=10\) - step1: Find the domain: \(\ln{\left(10\right)}+\ln{\left(x^{2}\right)}=10,x\neq 0\) - step2: Transform the expression: \(\ln{\left(10x^{2}\right)}=10\) - step3: Convert the logarithm into exponential form: \(10x^{2}=e^{10}\) - step4: Divide both sides: \(\frac{10x^{2}}{10}=\frac{e^{10}}{10}\) - step5: Divide the numbers: \(x^{2}=\frac{e^{10}}{10}\) - step6: Simplify the expression: \(x=\pm \sqrt{\frac{e^{10}}{10}}\) - step7: Simplify: \(x=\pm \frac{\sqrt{10}\times e^{5}}{10}\) - step8: Separate into possible cases: \(\begin{align}&x=\frac{\sqrt{10}\times e^{5}}{10}\\&x=-\frac{\sqrt{10}\times e^{5}}{10}\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&x=\frac{\sqrt{10}\times e^{5}}{10}\\&x=-\frac{\sqrt{10}\times e^{5}}{10}\end{align},x\neq 0\) - step10: Find the intersection: \(\begin{align}&x=\frac{\sqrt{10}\times e^{5}}{10}\\&x=-\frac{\sqrt{10}\times e^{5}}{10}\end{align}\) - step11: Rewrite: \(x_{1}=-\frac{\sqrt{10}\times e^{5}}{10},x_{2}=\frac{\sqrt{10}\times e^{5}}{10}\) The solutions to the equation \( \ln 10 + \ln x^2 = 10 \) are \( x = -\frac{\sqrt{10} \times e^5}{10} \) and \( x = \frac{\sqrt{10} \times e^5}{10} \).