(b) \( \int^{\text {an }} 9^{x}+3^{2 x+3}=43^{2 x} \) (d) \( 2^{x+3}-2^{x+2}=2^{x+2} \)

Let's solve each of the given exponential equations step by step. --- ### Problem (b) **Equation:** \[ 9^{x} + 3^{2x + 3} = 43^{2x} \] **Step 1: Express all terms with the same base if possible.** - Notice that \( 9 = 3^2 \), so \( 9^x = (3^2)^x = 3^{2x} \). - Similarly, \( 3^{2x + 3} = 3^{2x} \cdot 3^3 = 3^{2x} \cdot 27 \). **Step 2: Substitute these expressions back into the equation.** \[ 3^{2x} + 27 \cdot 3^{2x} = 43^{2x} \] \[ (1 + 27) \cdot 3^{2x} = 43^{2x} \] \[ 28 \cdot 3^{2x} = 43^{2x} \] **Step 3: Take the natural logarithm of both sides to solve for \( x \).** \[ \ln(28 \cdot 3^{2x}) = \ln(43^{2x}) \] \[ \ln(28) + \ln(3^{2x}) = \ln(43^{2x}) \] \[ \ln(28) + 2x \ln(3) = 2x \ln(43) \] **Step 4: Isolate \( x \).** \[ 2x \ln(43) - 2x \ln(3) = \ln(28) \] \[ 2x (\ln(43) - \ln(3)) = \ln(28) \] \[ x = \frac{\ln(28)}{2 (\ln(43) - \ln(3))} \] **Final Answer for (b):** \[ x = \frac{\ln(28)}{2 (\ln(43) - \ln(3))} \] --- ### Problem (d) **Equation:** \[ 2^{x + 3} - 2^{x + 2} = 2^{x + 2} \] **Step 1: Simplify each term.** - \( 2^{x + 3} = 2^3 \cdot 2^x = 8 \cdot 2^x \) - \( 2^{x + 2} = 2^2 \cdot 2^x = 4 \cdot 2^x \) **Step 2: Substitute these back into the equation.** \[ 8 \cdot 2^x - 4 \cdot 2^x = 4 \cdot 2^x \] \[ (8 - 4) \cdot 2^x = 4 \cdot 2^x \] \[ 4 \cdot 2^x = 4 \cdot 2^x \] **Step 3: Analyze the equation.** The equation simplifies to an identity: \[ 4 \cdot 2^x = 4 \cdot 2^x \] This is true for **all real numbers** \( x \). **Final Answer for (d):** All real numbers \( x \) satisfy the equation. ---