(6) \( 4^{x}+16=10 \cdot 2^{x} \)

To solve the equation \(4^{x} + 16 = 10 \cdot 2^{x}\), we can first rewrite \(4^x\) in terms of \(2^x\). Since \(4 = 2^2\), we have \(4^x = (2^2)^x = 2^{2x}\). Thus, we can rewrite the equation as follows: \[ 2^{2x} + 16 = 10 \cdot 2^{x} \] Next, we can let \(y = 2^x\). Therefore, \(2^{2x} = y^2\) and our equation transforms into: \[ y^2 + 16 = 10y \] Now, rearranging this gives us a standard form quadratic equation: \[ y^2 - 10y + 16 = 0 \] We can solve this quadratic equation using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -10\), and \(c = 16\): \[ y = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm \sqrt{36}}{2} \] Thus, we simplify further: \[ y = \frac{10 \pm 6}{2} \] This gives two potential solutions for \(y\): \[ y = \frac{16}{2} = 8 \quad \text{and} \quad y = \frac{4}{2} = 2 \] Now, recalling that \(y = 2^x\), we can solve for \(x\): 1. For \(y = 8\), we have \(2^x = 8\), which means \(x = 3\) because \(2^3 = 8\). 2. For \(y = 2\), we have \(2^x = 2\), which means \(x = 1\) because \(2^1 = 2\). Finally, the solutions to the equation \(4^{x}+16=10 \cdot 2^{x}\) are: \[ \boxed{1 \quad \text{and} \quad 3} \]