[PINNACLE COLLEGE RYNFIELD]Prove that: \[ \frac{14.7^{x-1}+7^{x+1}}{3^{-x}}=9 p \text { if } p=21 \]

We start with the expression \[ \frac{14 \cdot 7^{x-1} + 7^{x+1}}{3^{-x}}. \] **Step 1. Factor the numerator** Factor out the common factor \(7^{x-1}\) from the numerator: \[ 14 \cdot 7^{x-1} + 7^{x+1} = 7^{x-1}\left(14 + 7^2\right). \] Since \[ 7^2 = 49, \] the expression inside the parentheses becomes \[ 14 + 49 = 63. \] Thus, the numerator simplifies to \[ 7^{x-1} \cdot 63. \] **Step 2. Rewrite the entire expression** Substitute the factored numerator back into the expression: \[ \frac{63 \cdot 7^{x-1}}{3^{-x}}. \] Recall that dividing by \(3^{-x}\) is equivalent to multiplying by \(3^x\), so we have \[ 63 \cdot 7^{x-1} \cdot 3^x. \] **Step 3. Equate the expression to \(9p\) with \(p=21\)** We are given that \[ \frac{14 \cdot 7^{x-1} + 7^{x+1}}{3^{-x}} = 9p \quad \text{with} \quad p=21. \] Substitute \(p=21\): \[ 9p = 9 \times 21 = 189. \] Thus, our equation becomes \[ 63 \cdot 7^{x-1} \cdot 3^x = 189. \] **Step 4. Solve for \(x\)** Divide both sides of the equation by \(63\): \[ 7^{x-1} \cdot 3^x = \frac{189}{63} = 3. \] Notice that \(3\) can be written as \(3^1\) (and \(7^0\), if you wish): \[ 7^{x-1} \cdot 3^x = 7^0 \cdot 3^1. \] For the equality to hold, we require that the factors involving different bases match accordingly. - For the factor with base \(7\): Since \(7^{x-1} = 7^0\), we have \[ x - 1 = 0 \quad \Longrightarrow \quad x = 1. \] - For the factor with base \(3\): With \(x = 1\), we have \[ 3^x = 3^1, \] which is consistent. Thus, the unique solution is \[ x = 1. \] **Step 5. Verify the solution** Substitute \(x = 1\) back into the original expression: \[ \frac{14 \cdot 7^{1-1} + 7^{1+1}}{3^{-1}} = \frac{14 \cdot 7^0 + 7^2}{3^{-1}}. \] Since \(7^0 = 1\) and \(7^2 = 49\), the numerator becomes \[ 14 \times 1 + 49 = 14 + 49 = 63. \] The denominator is \[ 3^{-1} = \frac{1}{3}. \] Thus, the expression evaluates to \[ \frac{63}{\frac{1}{3}} = 63 \times 3 = 189. \] Since \[ 189 = 9 \times 21, \] we have shown that \[ \frac{14 \cdot 7^{x-1} + 7^{x+1}}{3^{-x}} = 9p \quad \text{with} \quad p=21, \] when \[ x = 1. \]