\( 5.2 ^ { x } + 3.2 ^ { x + 2 } = 68 \)

To solve the exponential equation: \[ 5.2^x + 3.2^{x + 2} = 68 \] Let's break it down step by step. ### Step 1: Simplify the Equation First, rewrite \(3.2^{x + 2}\) using the properties of exponents: \[ 3.2^{x + 2} = 3.2^x \times 3.2^2 = 3.2^x \times 10.24 \] Now, the equation becomes: \[ 5.2^x + 10.24 \times 3.2^x = 68 \] ### Step 2: Set Up for Numerical Solution This equation involves different exponential bases (\(5.2\) and \(3.2\)), making it challenging to solve algebraically. Therefore, we'll use numerical methods (such as the **Newton-Raphson method**) to approximate the value of \(x\). ### Step 3: Define the Function Let's define the function \(f(x)\) as: \[ f(x) = 5.2^x + 10.24 \times 3.2^x - 68 \] We need to find \(x\) such that \(f(x) = 0\). ### Step 4: Initial Estimates Let's test some initial values to narrow down the range where \(x\) lies. - **For \(x = 1\):** \[ f(1) = 5.2 + 10.24 \times 3.2 \approx 5.2 + 32.768 = 37.968 < 68 \] - **For \(x = 2\):** \[ f(2) = 5.2^2 + 10.24 \times 3.2^2 \approx 27.04 + 104.8576 = 131.8976 > 68 \] Thus, \(x\) lies between 1 and 2. ### Step 5: Refine the Estimate Continuing with more precise estimates: - **For \(x = 1.5\):** \[ f(1.5) \approx 11.8576 + 58.534 \approx 70.392 > 68 \] - **For \(x = 1.45\):** \[ f(1.45) \approx 10.887 + 55.29 \approx 66.177 < 68 \] Since \(f(1.45) < 68\) and \(f(1.5) > 68\), the solution lies between 1.45 and 1.5. Using linear approximation and iterative methods, we refine the estimate: \[ x \approx 1.4718 \] ### Step 6: Final Solution Thus, the solution to the equation \(5.2^x + 3.2^{x + 2} = 68\) is approximately: \[ x \approx 1.47 \] ### Verification To ensure accuracy, plug \(x \approx 1.47\) back into the original equation: \[ 5.2^{1.47} + 10.24 \times 3.2^{1.47} \approx 11.24 + 56.78 \approx 68.02 \approx 68 \] This confirms that our approximation is correct. **Final Answer:** The solution is approximately \( x \approx 1.47 \).