3 Solve in \( \mathbb{R} \) each of the following equations : 1) \( e^{3 x+1}=e^{2 x-2} \) 2) \( e^{x^{2}} \cdot e^{x}=1 \) 3) \( \frac{2 e^{x}+1}{e^{x}+1}>0 \) 4) \( e^{2 x}-e^{x}-6 \geq 0 \)

Solve the equation by following steps: - step0: Solve for \(x\): \(e^{3x+1}=e^{2x-2}\) - step1: Set the exponents equal: \(3x+1=2x-2\) - step2: Move the expression to the left side: \(3x-2x=-2-1\) - step3: Add and subtract: \(x=-2-1\) - step4: Add and subtract: \(x=-3\) Solve the equation \( e^{x^{2}} \times e^{x}=1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(e^{\left(x^{2}\right)}\times e^{x}=1\) - step1: Multiply: \(e^{x^{2}+x}=1\) - step2: Write in exponential form: \(e^{x^{2}+x}=e^{0}\) - step3: Set the exponents equal: \(x^{2}+x=0\) - step4: Factor the expression: \(x\left(x+1\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x=0\\&x+1=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=0\\&x=-1\end{align}\) - step7: Rewrite: \(x_{1}=-1,x_{2}=0\) Solve the equation \( \frac{2 e^{x}+1}{e^{x}+1}>0 \). Solve the inequality by following steps: - step0: Solve for \(x\): \(\frac{2e^{x}+1}{e^{x}+1}>0\) - step1: The statement is true for any value of \(x:\) \(x \in \mathbb{R}\) Solve the equation \( e^{2 x}-e^{x}-6 \geq 0 \). Solve the inequality by following steps: - step0: Solve for \(x\): \(e^{2x}-e^{x}-6\geq 0\) - step1: Factor the expression: \(\left(e^{x}-3\right)\left(e^{x}+2\right)\geq 0\) - step2: Separate into possible cases: \(\begin{align}&\left\{ \begin{array}{l}e^{x}-3\geq 0\\e^{x}+2\geq 0\end{array}\right.\\&\left\{ \begin{array}{l}e^{x}-3\leq 0\\e^{x}+2\leq 0\end{array}\right.\end{align}\) - step3: Solve the inequality: \(\begin{align}&\left\{ \begin{array}{l}x\geq \ln{\left(3\right)}\\x \in \mathbb{R}\end{array}\right.\\&\left\{ \begin{array}{l}x\leq \ln{\left(3\right)}\\x \notin \mathbb{R}\end{array}\right.\end{align}\) - step4: Find the intersection: \(\begin{align}&x\geq \ln{\left(3\right)}\\&x \notin \mathbb{R}\end{align}\) - step5: Find the union: \(x\geq \ln{\left(3\right)}\) Let's solve each of the equations step by step: 1) **Equation:** \( e^{3x+1} = e^{2x-2} \) To solve this equation, we can take the natural logarithm of both sides: \[ 3x + 1 = 2x - 2 \] Rearranging gives: \[ 3x - 2x = -2 - 1 \implies x = -3 \] **Solution:** \( x = -3 \) 2) **Equation:** \( e^{x^2} \cdot e^{x} = 1 \) We can combine the exponents: \[ e^{x^2 + x} = 1 \] Taking the natural logarithm of both sides: \[ x^2 + x = 0 \] Factoring gives: \[ x(x + 1) = 0 \] Thus, the solutions are: \[ x_1 = 0, \quad x_2 = -1 \] **Solution:** \( x = -1, 0 \) 3) **Inequality:** \( \frac{2e^{x} + 1}{e^{x} + 1} > 0 \) The expression is positive when the numerator is positive (since the denominator is always positive for all \( x \)): \[ 2e^{x} + 1 > 0 \] This is true for all \( x \) because \( e^{x} > 0 \) for all real \( x \). **Solution:** \( x \in \mathbb{R} \) (All real solutions) 4) **Inequality:** \( e^{2x} - e^{x} - 6 \geq 0 \) Let \( y = e^{x} \). Then the inequality becomes: \[ y^2 - y - 6 \geq 0 \] Factoring gives: \[ (y - 3)(y + 2) \geq 0 \] The critical points are \( y = 3 \) and \( y = -2 \). Since \( y = e^{x} > 0 \), we only consider \( y \geq 3 \): \[ e^{x} \geq 3 \implies x \geq \ln(3) \] **Solution:** \( x \geq \ln(3) \) In summary, the solutions are: 1. \( x = -3 \) 2. \( x = -1, 0 \) 3. \( x \in \mathbb{R} \) 4. \( x \geq \ln(3) \)