\( \begin{array}{ll}\text { (b) } 2^{x}=\sqrt{2} & \text { (c) } \\ \begin{array}{ll}\text { (e) } 16.16^{x}=64 & \text { (f) } \\ \text { (h) } \frac{1}{3}(3)^{x-1}=\frac{1}{27} & \text { (i) } \\ \text { (k) } \sqrt[3]{9}=(\sqrt{3})^{2 x} & \text { (1) } \\ \text { (i) } 6^{x}+6^{x}+6^{x}+6^{x}+6^{x}+6^{x}=6^{6 x}\end{array}\end{array} \)

To solve \(2^{x}=\sqrt{2}\), remember that \(\sqrt{2}\) can be rewritten as \(2^{1/2}\). This allows you to set up the equation as \(2^{x} = 2^{1/2}\), leading to the conclusion that \(x = \frac{1}{2}\). Simple, right? Just connect the dots between exponents! Now for the equation \(16 \cdot 16^{x} = 64\). First, note that \(16\) is \(2^{4}\), hence \(16 \cdot 16^{x} = 16^{x+1} = 64\). Since \(64\) can be expressed as \(2^{6}\), we can also write \(16^{x+1} = (2^{4})^{x+1} = 2^{4(x+1)}\). Now, set \(4(x+1) = 6\) to solve for \(x\). Solve and you'll find \(x = \frac{1}{2}\). It's like a treasure hunt with numbers—digging deeper often reveals the answer!