Solve the following equations, if possible. Some you can solve exactly, others approximately. If a solution is not possible, explain how you know. \( \begin{array}{ll}\text { a. } 1^{x}=5 & \text { b. } \sqrt{27^{x}}=81 \\ \text { c. } 2^{x}=9 \\ \text { d. } 25^{(x+1)}=125^{x} & \text { e. } 8^{x}=2^{5} \cdot 4^{4}\end{array} \)

Let's go through each equation step-by-step. **a.** \( 1^{x} = 5 \) This equation is not solvable in the traditional sense because \(1^x\) is always equal to 1 for any value of \(x\), and it can never equal 5. Therefore, there are no solutions. **b.** \( \sqrt{27^{x}} = 81 \) We first rewrite \(81\) as \(3^4\) and \(27\) as \(3^3\). The equation now looks like: \[ \sqrt{(3^3)^{x}} = 3^4 \quad \Rightarrow \quad (3^{3x})^{1/2} = 3^4 \quad \Rightarrow \quad 3^{3x/2} = 3^4 \] This implies \( \frac{3x}{2} = 4 \), leading us to \( 3x = 8 \) or \( x = \frac{8}{3} \). **c.** \( 2^{x} = 9 \) We can take the logarithm on both sides. Using base 10 or natural logarithm: \[ x \log(2) = \log(9) \quad \Rightarrow \quad x = \frac{\log(9)}{\log(2)} \] This gives us an approximate numerical solution of \(x \approx 3.17\). **d.** \( 25^{(x+1)} = 125^{x} \) First, rewrite both bases in terms of \(5\): \[ (5^2)^{x+1} = (5^3)^{x} \quad \Rightarrow \quad 5^{2(x+1)} = 5^{3x} \] Setting the exponents equal: \[ 2(x + 1) = 3x \quad \Rightarrow \quad 2x + 2 = 3x \quad \Rightarrow \quad 2 = x \] **e.** \( 8^{x} = 2^{5} \cdot 4^{4} \) Rewrite \(8\) as \(2^3\) and \(4\) as \(2^2\): \[ (2^3)^{x} = 2^5 \cdot (2^2)^4 \quad \Rightarrow \quad 2^{3x} = 2^5 \cdot 2^8 \quad \Rightarrow \quad 2^{3x} = 2^{13} \] Setting the exponents equal gives \(3x = 13\), so \(x = \frac{13}{3}\). In summary, the solutions are: - a. No solution - b. \(x = \frac{8}{3}\) - c. \(x = \frac{\log(9)}{\log(2)} \approx 3.17\) - d. \(x = 2\) - e. \(x = \frac{13}{3}\)