\( 6.10 ^ { x } + 10 ^ { 2 x + 2 } - 106 = 0 \)

To solve the equation: \[ 6 \cdot 10^{x} + 10^{2x + 2} - 106 = 0 \] Let's proceed step-by-step. ### Step 1: Simplify the Equation First, recognize that \(10^{2x + 2}\) can be rewritten using the properties of exponents: \[ 10^{2x + 2} = 10^{2x} \cdot 10^{2} = 100 \cdot 10^{2x} \] So, the equation becomes: \[ 6 \cdot 10^{x} + 100 \cdot 10^{2x} - 106 = 0 \] ### Step 2: Substitute to Simplify Let’s make a substitution to simplify the equation. Let: \[ y = 10^{x} \] Then: \[ 10^{2x} = (10^{x})^{2} = y^{2} \] Substituting into the original equation: \[ 6y + 100y^{2} - 106 = 0 \] ### Step 3: Form a Quadratic Equation This is a quadratic equation in terms of \(y\): \[ 100y^{2} + 6y - 106 = 0 \] To simplify, divide the entire equation by 2: \[ 50y^{2} + 3y - 53 = 0 \] ### Step 4: Solve the Quadratic Equation Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a = 50\), \(b = 3\), and \(c = -53\): \[ \Delta = b^{2} - 4ac = 3^{2} - 4 \cdot 50 \cdot (-53) = 9 + 10600 = 10609 \] \[ \sqrt{\Delta} = \sqrt{10609} = 103 \] So, \[ y = \frac{-3 \pm 103}{100} \] This gives two potential solutions: \[ y_1 = \frac{-3 + 103}{100} = \frac{100}{100} = 1 \] \[ y_2 = \frac{-3 - 103}{100} = \frac{-106}{100} = -1.06 \] ### Step 5: Back-Substitute for \(x\) Recall that \(y = 10^{x}\). Since \(10^{x}\) is always positive, \(y = -1.06\) is not a valid solution. Thus, we only consider \(y = 1\): \[ 10^{x} = 1 \] Taking the logarithm (base 10) of both sides: \[ x = \log_{10}(1) = 0 \] ### Conclusion The only solution to the equation is: \[ x = 0 \] ### Verification Plugging \(x = 0\) back into the original equation: \[ 6 \cdot 10^{0} + 10^{2(0) + 2} - 106 = 6 \cdot 1 + 10^{2} - 106 = 6 + 100 - 106 = 0 \] The equation holds true, confirming that \(x = 0\) is the correct solution. **Answer:** The solution is \( x = 0 \).