Exercice 2 Résoudre suivantes: \( \begin{array}{l}\text { a) } \ln x=3 \\ \text { d) }[\ln x][\ln (x-1)]=0 \text { e } \\ \begin{array}{l}\text { f) } 2 \ln x=\ln (2 x-1) \\ \text { g) } \\ \text { h) } \ln \left(x^{2}-x\right)=\ln 2+\ln 3 \\ \text { j) } \ln (x+4)+\ln (2 x-2)- \\ \text { k) } \ln (x+3)+\ln (x+2)=1 \\ \text { I) }(\ln x)^{2}-2 \ln x+1=0 \text { m) } \\ \text { Exercice } 3 \text { Résoudre }\end{array}\end{array} \).

Solve the equation by following steps: - step0: Solve for \(x\): \(2\ln{\left(x\right)}=\ln{\left(2x-1\right)}\) - step1: Find the domain: \(2\ln{\left(x\right)}=\ln{\left(2x-1\right)},x>\frac{1}{2}\) - step2: Move the expression to the left side: \(2\ln{\left(x\right)}-\ln{\left(2x-1\right)}=0\) - step3: Add the terms: \(\ln{\left(\frac{x^{2}}{2x-1}\right)}=0\) - step4: Convert the logarithm into exponential form: \(\frac{x^{2}}{2x-1}=e^{0}\) - step5: Evaluate the power: \(\frac{x^{2}}{2x-1}=1\) - step6: Cross multiply: \(x^{2}=2x-1\) - step7: Move the expression to the left side: \(x^{2}-\left(2x-1\right)=0\) - step8: Remove the parentheses: \(x^{2}-2x+1=0\) - step9: Factor the expression: \(\left(x-1\right)^{2}=0\) - step10: Simplify the expression: \(x-1=0\) - step11: Move the constant to the right side: \(x=0+1\) - step12: Remove 0: \(x=1\) - step13: Check if the solution is in the defined range: \(x=1,x>\frac{1}{2}\) - step14: Find the intersection: \(x=1\) Solve the equation \( (\ln x)^{2}-2 \ln x+1=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left(\ln{\left(x\right)}\right)^{2}-2\ln{\left(x\right)}+1=0\) - step1: Find the domain: \(\left(\ln{\left(x\right)}\right)^{2}-2\ln{\left(x\right)}+1=0,x>0\) - step2: Solve using substitution: \(t^{2}-2t+1=0\) - step3: Factor the expression: \(\left(t-1\right)^{2}=0\) - step4: Simplify the expression: \(t-1=0\) - step5: Move the constant to the right side: \(t=0+1\) - step6: Remove 0: \(t=1\) - step7: Substitute back: \(\ln{\left(x\right)}=1\) - step8: Convert the logarithm into exponential form: \(x=e^{1}\) - step9: Evaluate the power: \(x=e\) - step10: Check if the solution is in the defined range: \(x=e,x>0\) - step11: Find the intersection: \(x=e\) Solve the equation \( \ln (x+4)+\ln (2 x-2)=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\ln{\left(x+4\right)}+\ln{\left(2x-2\right)}=0\) - step1: Find the domain: \(\ln{\left(x+4\right)}+\ln{\left(2x-2\right)}=0,x>1\) - step2: Transform the expression: \(\ln{\left(\left(x+4\right)\left(2x-2\right)\right)}=0\) - step3: Convert the logarithm into exponential form: \(\left(x+4\right)\left(2x-2\right)=e^{0}\) - step4: Evaluate the power: \(\left(x+4\right)\left(2x-2\right)=1\) - step5: Expand the expression: \(2x^{2}+6x-8=1\) - step6: Move the expression to the left side: \(2x^{2}+6x-8-1=0\) - step7: Subtract the numbers: \(2x^{2}+6x-9=0\) - step8: Solve using the quadratic formula: \(x=\frac{-6\pm \sqrt{6^{2}-4\times 2\left(-9\right)}}{2\times 2}\) - step9: Simplify the expression: \(x=\frac{-6\pm \sqrt{6^{2}-4\times 2\left(-9\right)}}{4}\) - step10: Simplify the expression: \(x=\frac{-6\pm \sqrt{108}}{4}\) - step11: Simplify the expression: \(x=\frac{-6\pm 6\sqrt{3}}{4}\) - step12: Separate into possible cases: \(\begin{align}&x=\frac{-6+6\sqrt{3}}{4}\\&x=\frac{-6-6\sqrt{3}}{4}\end{align}\) - step13: Simplify the expression: \(\begin{align}&x=\frac{-3+3\sqrt{3}}{2}\\&x=\frac{-6-6\sqrt{3}}{4}\end{align}\) - step14: Simplify the expression: \(\begin{align}&x=\frac{-3+3\sqrt{3}}{2}\\&x=-\frac{3+3\sqrt{3}}{2}\end{align}\) - step15: Check if the solution is in the defined range: \(\begin{align}&x=\frac{-3+3\sqrt{3}}{2}\\&x=-\frac{3+3\sqrt{3}}{2}\end{align},x>1\) - step16: Find the intersection: \(x=\frac{-3+3\sqrt{3}}{2}\) Solve the equation \( [\ln x][\ln (x-1)]=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left(\ln{\left(x\right)}\right)\left(\ln{\left(x-1\right)}\right)=0\) - step1: Find the domain: \(\left(\ln{\left(x\right)}\right)\left(\ln{\left(x-1\right)}\right)=0,x>1\) - step2: Simplify: \(\ln{\left(x\right)}\times \ln{\left(x-1\right)}=0\) - step3: Separate into possible cases: \(\begin{align}&\ln{\left(x\right)}=0\\&\ln{\left(x-1\right)}=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=1\\&x=2\end{align}\) - step5: Check if the solution is in the defined range: \(\begin{align}&x=1\\&x=2\end{align},x>1\) - step6: Find the intersection: \(x=2\) Solve the equation \( \ln (x+3)+\ln (x+2)=1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\ln{\left(x+3\right)}+\ln{\left(x+2\right)}=1\) - step1: Find the domain: \(\ln{\left(x+3\right)}+\ln{\left(x+2\right)}=1,x>-2\) - step2: Transform the expression: \(\ln{\left(\left(x+3\right)\left(x+2\right)\right)}=1\) - step3: Convert the logarithm into exponential form: \(\left(x+3\right)\left(x+2\right)=e^{1}\) - step4: Evaluate the power: \(\left(x+3\right)\left(x+2\right)=e\) - step5: Expand the expression: \(x^{2}+5x+6=e\) - step6: Move the expression to the left side: \(x^{2}+5x+6-e=0\) - step7: Solve using the quadratic formula: \(x=\frac{-5\pm \sqrt{5^{2}-4\left(6-e\right)}}{2}\) - step8: Simplify the expression: \(x=\frac{-5\pm \sqrt{1+4e}}{2}\) - step9: Separate into possible cases: \(\begin{align}&x=\frac{-5+\sqrt{1+4e}}{2}\\&x=\frac{-5-\sqrt{1+4e}}{2}\end{align}\) - step10: Rewrite the fraction: \(\begin{align}&x=\frac{-5+\sqrt{1+4e}}{2}\\&x=-\frac{5+\sqrt{1+4e}}{2}\end{align}\) - step11: Check if the solution is in the defined range: \(\begin{align}&x=\frac{-5+\sqrt{1+4e}}{2}\\&x=-\frac{5+\sqrt{1+4e}}{2}\end{align},x>-2\) - step12: Find the intersection: \(x=\frac{-5+\sqrt{1+4e}}{2}\) Solve the equation \( \ln \left(x^{2}-x\right)=\ln 2+\ln 3 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\ln{\left(x^{2}-x\right)}=\ln{\left(2\right)}+\ln{\left(3\right)}\) - step1: Find the domain: \(\ln{\left(x^{2}-x\right)}=\ln{\left(2\right)}+\ln{\left(3\right)},x \in \left(-\infty,0\right)\cup \left(1,+\infty\right)\) - step2: Add the terms: \(\ln{\left(x^{2}-x\right)}=\ln{\left(6\right)}\) - step3: Evaluate the logarithm: \(x^{2}-x=6\) - step4: Move the expression to the left side: \(x^{2}-x-6=0\) - step5: Factor the expression: \(\left(x-3\right)\left(x+2\right)=0\) - step6: Separate into possible cases: \(\begin{align}&x-3=0\\&x+2=0\end{align}\) - step7: Solve the equation: \(\begin{align}&x=3\\&x=-2\end{align}\) - step8: Check if the solution is in the defined range: \(\begin{align}&x=3\\&x=-2\end{align},x \in \left(-\infty,0\right)\cup \left(1,+\infty\right)\) - step9: Find the intersection: \(\begin{align}&x=3\\&x=-2\end{align}\) - step10: Rewrite: \(x_{1}=-2,x_{2}=3\) Voici les solutions pour chaque équation : ### a) \( \ln x = 3 \) Pour résoudre cette équation, nous exponentions les deux côtés : \[ x = e^3 \] ### d) \( [\ln x][\ln (x-1)] = 0 \) Cette équation est satisfaite si l'un des facteurs est égal à zéro : 1. \( \ln x = 0 \) \(\Rightarrow x = 1\) 2. \( \ln (x-1) = 0 \) \(\Rightarrow x - 1 = 1 \Rightarrow x = 2\) Donc, les solutions sont : \[ x = 1 \quad \text{ou} \quad x = 2 \] ### f) \( 2 \ln x = \ln (2x - 1) \) En utilisant les propriétés des logarithmes, nous pouvons réécrire l'équation : \[ \ln x^2 = \ln (2x - 1) \Rightarrow x^2 = 2x - 1 \] En réarrangeant, nous avons : \[ x^2 - 2x + 1 = 0 \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1 \] ### h) \( \ln (x^2 - x) = \ln 2 + \ln 3 \) En utilisant les propriétés des logarithmes, nous avons : \[ \ln (x^2 - x) = \ln (6) \Rightarrow x^2 - x = 6 \] En réarrangeant, nous avons : \[ x^2 - x - 6 = 0 \] En factorisant, nous obtenons : \[ (x - 3)(x + 2) = 0 \Rightarrow x_1 = 3, \quad x_2 = -2 \] ### j) \( \ln (x + 4) + \ln (2x - 2) = 0 \) En utilisant les propriétés des logarithmes, nous avons : \[ \ln ((x + 4)(2x - 2)) = 0 \Rightarrow (x + 4)(2x - 2) = 1 \] En résolvant cette équation, nous obtenons : \[ x = \frac{-3 + 3\sqrt{3}}{2} \] ### k) \( \ln (x + 3) + \ln (x + 2) = 1 \) En utilisant les propriétés des logarithmes, nous avons : \[ \ln ((x + 3)(x + 2)) = 1 \Rightarrow (x + 3)(x + 2) = e \] En résolvant cette équation, nous obtenons : \[ x = \frac{-5 + \sqrt{1 + 4e}}{2} \] ### l) \( (\ln x)^2 - 2 \ln x + 1 = 0 \) Cette équation peut être factorisée : \[ (\ln x - 1)^2 = 0 \Rightarrow \ln x = 1 \Rightarrow x = e \] ### Résumé des solutions : - a) \( x = e^3 \) - d) \( x = 1 \) ou \( x = 2 \) - f) \( x = 1 \) - h) \( x = 3 \) ou \( x = -2 \) - j) \( x = \frac{-3 + 3\sqrt{3}}{2} \) - k) \( x = \frac{-5 + \sqrt{1 + 4e}}{2} \) - l) \( x = e \)