Solve for \( x \) : (a) \( 4.3^{x}+3^{x}=15 \) (c) \( 2^{x+2}+2^{x-2}+2^{x}=84 \)

To solve the equation \( 4.3^{x}+3^{x}=15 \), we can try different values of \( x \) to see if we can find a solution. Let's test \( x = 1 \): \[ 4.3^{1} + 3^{1} = 4 \cdot 3 + 3 = 12 + 3 = 15 \] So, \( x = 1 \) is a solution. For the second equation \( 2^{x+2}+2^{x-2}+2^{x}=84 \), we can first rewrite the equation in a more manageable form. Notice that: \[ 2^{x+2} = 4 \cdot 2^{x}, \quad 2^{x-2} = \frac{2^{x}}{4} \] Substituting these back into the equation gives: \[ 4 \cdot 2^{x} + \frac{2^{x}}{4} + 2^{x} = 84 \] Let \( y = 2^{x} \), so we rewrite the equation: \[ 4y + \frac{y}{4} + y = 84 \] Multiplying everything by 4 to eliminate the fraction: \[ 16y + y + 4y = 336 \] \[ 21y = 336 \] \[ y = \frac{336}{21} = 16 \] Now substituting back, since \( y = 2^{x} \), we get: \[ 2^{x} = 16 \implies 2^{x} = 2^{4} \implies x = 4 \] So, the solutions are \( x = 1 \) for the first equation and \( x = 4 \) for the second equation.