EXERCISE 7 Solve for \( x \) : \( \begin{array}{lll}\text { (a) } 3^{x}=1 & \text { (b) } 2^{x}=\sqrt{2} & \text { (c) } 4^{x}=8 \\ \text { (d) }\left(\frac{1}{9}\right)^{x}=27 & \text { (e) } 16 \cdot 16^{x}=64 & \text { (f) } 3^{x+1}=\frac{\sqrt{n}}{3} \\ \text { (g) } 4\left(\frac{1}{4}\right)^{x-1}=8 & \text { (h) } \frac{1}{3}(3)^{x-1}=\frac{1}{27} & \text { (i) } 8^{x} \cdot 16^{x-1} \\ \text { (i) } \frac{3^{2 x-1}}{3^{x}}=3 & \text { (k) } \sqrt[3]{9}=(\sqrt{3})^{2 x} & \text { (l) } \\ \text { (m) } 8^{-x}-2.4^{x-1}=0 & \text { (n) } 6^{x}+6^{x}+6^{x}+6^{x}+6^{x}+6^{x}=6^{6 x}\end{array} \)

Certainly! Let’s dive into solving these equations step by step! (a) To solve \(3^{x} = 1\), we recognize that \(1 = 3^0\). Therefore, \(x = 0\). (b) In \(2^{x} = \sqrt{2}\), we rewrite \(\sqrt{2}\) as \(2^{1/2}\). Thus, \(x = \frac{1}{2}\). (c) For \(4^{x} = 8\), we can express \(8\) as \(2^3\). Since \(4 = 2^2\), we have \(2^{2x} = 2^3\), leading to \(2x = 3\) and \(x = \frac{3}{2}\). (d) In \(\left(\frac{1}{9}\right)^{x} = 27\), we convert \(\frac{1}{9}\) to \(3^{-2}\) and \(27\) to \(3^3\). Thus, we have \(3^{-2x} = 3^3\), resulting in \(-2x = 3\) and \(x = -\frac{3}{2}\). (e) For \(16 \cdot 16^{x}=64\), we rewrite \(64\) as \(16^{3/2}\) (since \(64 = 4^3\) and \(4 = 16^{1/2}\)). This gives us \(16^{1 + x} = 16^{3/2}\), so \(1 + x = \frac{3}{2}\), leading to \(x = \frac{1}{2}\). (f) From \(3^{x+1} = \frac{\sqrt{n}}{3}\), we can express \(\frac{\sqrt{n}}{3}\) as \(3^{-1}n^{1/2}\) and simplify to \(3^{x+1} = n^{1/2} \cdot 3^{-1}\) which results in \(3^{x+2} = n^{1/2}\). (g) In the equation \(4\left(\frac{1}{4}\right)^{x-1} = 8\), simplifying gives \(4 \cdot 4^{-(x-1)} = 8\), or \(4^{2-x} = 8\) is equivalent to \(2^{4-2x} = 2^3\) leading us to \(4 - 2x = 3\), thus \(x = \frac{1}{2}\). (h) With \(\frac{1}{3}(3)^{x-1}=\frac{1}{27}\), we multiply both sides by 3, yielding \(3^{x-1} = \frac{1}{9} = 3^{-2}\). As a result, \(x-1 = -2\) which means \(x = -1\). (i) In \(8^{x} \cdot 16^{x-1}\), we can convert that into \(2^{3x} \cdot 2^{4(x-1)} = 2^{3x + 4(x-1)} = 2^{7x-4}\). (j) The equation \(\frac{3^{2x-1}}{3^{x}}=3\) simplifies to \(3^{x-1} = 3^1\), yielding \(x - 1 = 1\) hence \(x = 2\). (k) To solve \(\sqrt[3]{9}=(\sqrt{3})^{2x}\), we can write \(9\) as \(3^{2}\). So, \(3^{2/3} = 3^{x}\) gives \(x = \frac{2}{3}\). (l) In \(8^{-x}-2\cdot4^{x-1}=0\), first change \(4\) to \(2^2\) so this becomes \(2^{-3x} - 2\cdot2^{2(x-1)} = 0\). This can lead to a resolution by equating the exponents. (m) The equation \(6^{x}+6^{x}+6^{x}+6^{x}+6^{x}+6^{x}=6^{6x} \) combines terms yielding \(6 \cdot 6^{x} = 6^{6x}\), or \(6^{x+1}=6^{6x}\). So, \(x + 1 = 6x\) leading to \(x = \frac{1}{5}\). I