1. Solve for \( x \) : \( \begin{array}{ll}\text { (a) } 4.3^{x}+3^{x}=15 & \text { (b) } 2^{x+1}+2^{x-1}=5 \\ \text { (c) } 2^{x+2}+2^{x-2}+2^{x}=84 & \text { (d) } 3\left(3^{x+2}+3^{x-1}\right)=84\end{array} \)
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For equation (a) \( 4.3^{x} + 3^{x} = 15 \), we can make a substitution \( y = 3^x \), so \( 4y + y = 15 \), which simplifies to \( 5y = 15 \) and thus \( y = 3 \). By substituting back, we find \( 3^x = 3 \), leading to \( x = 1 \). For equation (b) \( 2^{x+1} + 2^{x-1} = 5 \), we can rewrite it as \( 2 \cdot 2^x + \frac{2^x}{2} = 5 \) or \( 2^{x+1} + 2^{x-1} = 5 \rightarrow 2^{x} (2 + \frac{1}{2}) = 5 \). This results in \( 2^{x} \cdot \frac{5}{2} = 5 \), and solving gives \( 2^{x} = 2 \), so \( x = 1 \) as well. For (c) \( 2^{x+2} + 2^{x-2} + 2^{x} = 84 \), rewrite each term: \( 4 \cdot 2^{x} + \frac{1}{4} \cdot 2^{x} + 2^{x} = 84 \). Combine these to find \( (4 + \frac{1}{4} + 1) 2^{x} = 84 \), simplifying further gives \( \frac{21}{4} 2^{x} = 84 \) leading to \( 2^{x} = 16 \), hence \( x = 4 \). For (d) \( 3(3^{x+2} + 3^{x-1}) = 84 \), we simplify it to \( 3^{x+2} + 3^{x-1} = 28 \). Making a substitution \( z = 3^x \), we have \( 9z + \frac{z}{3} = 28 \) or \( (9 + \frac{1}{3})z = 28 \), which simplifies to \( \frac{28}{\frac{28}{3}} \rightarrow z = 3 \), yielding \( x = 1 \). In summary: (a) \( x = 1 \) (b) \( x = 1 \) (c) \( x = 4 \) (d) \( x = 1 \)
