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\( \begin{array}{ll}\text { a) } 1+2^{1-2 x}=4,5 \times 2^{-x} & \text { b) } 3^{-2 x}+8.3^{-x}=9 \\ \text { c) } 5^{x}+5=2.5^{2-x} & \text { d) } 2^{x}=2^{4-x} \\ \text { e) } 10^{x}-11+10^{1-x}=0 & \text { f) } 3^{2-x}+2^{3}=3^{x}\end{array} \)

Ask by Marsh Spencer. in South Africa
Feb 03,2025

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Here are the solutions for each equation: a) \( x = -1 \) or \( x = 2 \) b) \( x = 0 \) c) \( x = 1 \) d) \( x = 2 \) e) \( x = 0 \) or \( x = 1 \) f) \( x = 2 \)

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To solve the given equations, a good practice is to convert them into forms that can be tackled easily. For instance, in part (a), rewrite \( 2^{1-2x} \) using properties of exponents to isolate \( x \). Meanwhile, in part (d), notice that since the equation is balanced with same bases, you can set the exponents equal to each other, which significantly simplifies the problem! Another clever tip is to always check for extraneous solutions, especially when you manipulate equations involving exponents. For example, when working with logarithms or squaring both sides of an equation, there could be solutions that don’t satisfy the original equation. Always plug your solutions back to verify!

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