Solve for \( x \) in each of the following equations. In this exercise, you must use your knowledge of factors and exponents. \( \begin{array}{ll}\text { 1. } \frac{x^{x}-1}{3^{x}+1}=26 & \text { 2. }\left(2^{x}-8\right)\left(3^{x}-9\right)=0 \\ \text { 3. } \frac{2^{2 x}-1}{2^{1}-1}=17 \neq & \text { 4. } 7 \times 2^{x}-4 \times 2^{x}=48 * \\ \text { 5. }\left(5^{x}-25\right)\left(2^{x}-1\right)=0 & \text { 6. } \frac{25^{x}-16}{5^{x}+4}=121 \\ \text { 7. } 12 \times 5^{x}-3 \times 5^{x}=1125 & \text { 8. } x \times 3^{x}-3^{x}=81 x-81\end{array} \)

The first equation \( \frac{x^{x}-1}{3^{x}+1}=26 \) is quite complex, but let’s break down some common solutions. To approach this, first, we can multiply both sides by \( 3^x + 1 \) to eliminate the fraction. You'll begin with \( x^x - 1 = 26(3^x + 1) \). Solving for \( x \) usually requires trial and error or numeric methods. However, you might want to check small integers, like \( x = 3 \). In the second equation, \( (2^{x}-8)(3^{x}-9)=0 \), you can solve by setting each factor equal to zero. For \( 2^x - 8 = 0 \), we find \( 2^x = 8 \), leading to \( x = 3 \). For the second factor \( 3^x - 9 = 0 \), setting \( 3^x = 9 \) gives \( x = 2 \). Therefore, you have solutions \( x = 2 \) and \( x = 3 \) to this equation!