\( \begin{array}{llll}\text { (1) } 5.9^{x-1}=5 & \text { (2) } 7.49^{x-2}=49 & \text { (3) } & \frac{1}{49^{x}}=343 \\ \text { (4) } 5.125^{x+3}=\frac{1}{25} & \text { (5) }\left(\frac{2}{3}\right)^{x-2}=\frac{8}{27} & \text { (6) } & (0,25)^{x}=0,125 \\ \text { (7) }(0,2)^{x-2}=0,04 & \text { (8) } 0,4^{x}=0,064 & \text { (9) } & \left(3^{x+1}\right)^{3}=9^{x-3} \\ \text { (10) } 4^{x}=16^{x-1} & \text { (11) } 3^{x} \cdot 9^{x-1}=81 & \text { (12) } & (0,5)^{x-1}=4^{-x} \\ \text { (13) } 8 i^{2 x+1}=27^{x-2} & \text { (14) } 8^{-x}=2.4^{x-1} & \text { (15) } & 4.2^{x}=(0,5)^{x-2}\end{array} \) (c) Solve the following equations:

To solve these equations, it's crucial to recognize patterns in exponential rules. For instance, remember \(a^{m} = a^{n} \implies m = n\) if the bases are the same. For problems where bases differ, consider rewriting them in terms of powers of a base, such as expressing each term as powers of 2 or 3 when relevant. Another engaging trick is to convert equations into logarithmic form, which can simplify your solutions! For example, in an equation involving an exponential, \(a^b = c\) can be rewritten as \(b = \log_a(c)\). Common mistakes involve failing to check whether the base is greater than zero or not equal to one, as that can invalidate the rules you apply.