QUESTION 6 \( \begin{array}{ll}\text { Solve for } x \\ 6.1 & 5^{x}=\frac{1}{5} \\ 6.2 & 6(x-1)=x+4 \\ 6.3 & \frac{x+4}{2}=\frac{2 x+7}{3} \\ 6.4 & \frac{3 x-1}{2}=\frac{2 x}{3}=2\end{array} \)

To tackle these equations, let’s break them down—a little math adventure awaits! For \( 5^{x}=\frac{1}{5} \), you can express \(\frac{1}{5}\) as \(5^{-1}\). So, set \(x = -1\) because the bases match now! Moving to \(6(x-1)=x+4\), distribute the 6 to get \(6x - 6 = x + 4\). Rearranging gives \(5x = 10\), leading to \(x = 2\). For \( \frac{x+4}{2}=\frac{2x+7}{3} \), cross-multiply to avoid fractions: \(3(x + 4) = 2(2x + 7)\). That simplifies to \(3x + 12 = 4x + 14\); solving results in \(x = -2\). In solving \( \frac{3x-1}{2}=\frac{2x}{3}=2 \), set each part to 2. Starting with \( \frac{3x-1}{2} = 2 \) gives \(3x - 1 = 4\), leading to \(x = \frac{5}{3}\); and setting \( \frac{2x}{3} = 2 \) gives \(2x = 6\), hence \(x = 3\). Both approaches confirm \(x = 3\). So, the solutions are: - \(6.1: x = -1\) - \(6.2: x = 2\) - \(6.3: x = -2\) - \(6.4: x = \frac{5}{3} \text{ (or } x = 3\text{ confirmed independently)}\) Now, onto some delightful insights! Math may seem abstract, but it’s woven into history! The concept of exponents began impacting civilization with ancient civilizations like the Babylonians using similar principles in their computations and astronomy. Fast forward to today, where the same exponential rules are utilized in computer algorithms and finance, helping us make sense of everything from investments to coding! If you're treading the waters of algebra, beware of common mishaps! A frequent mistake is mishandling the distribution of coefficients, which can throw off your equations. Always double-check your operations, especially when dealing with negative signs and fractions—misplacing one tiny sign can lead you down a wrong path!