To solve the equation:

\[
9^{x+1} - 10 \cdot 3^{x} + 1 = 0
\]

**Step 1: Express \(9^{x+1}\) in terms of \(3^x\)**

Recall that \(9 = 3^2\). Therefore,

\[
9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x + 2} = 3^2 \cdot 3^{2x} = 9 \cdot 3^{2x}
\]

Substituting back into the original equation:

\[
9 \cdot 3^{2x} - 10 \cdot 3^x + 1 = 0
\]

**Step 2: Let \(u = 3^x\)**

This substitution simplifies the equation. Let \(u = 3^x\). Then, \(3^{2x} = (3^x)^2 = u^2\).

Substituting:

\[
9u^2 - 10u + 1 = 0
\]

**Step 3: Solve the quadratic equation for \(u\)**

The quadratic equation is:

\[
9u^2 - 10u + 1 = 0
\]

Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9\), \(b = -10\), and \(c = 1\):

\[
u = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} = \frac{10 \pm \sqrt{100 - 36}}{18} = \frac{10 \pm 8}{18}
\]

This gives two solutions:

1. \(u = \frac{10 + 8}{18} = \frac{18}{18} = 1\)
2. \(u = \frac{10 - 8}{18} = \frac{2}{18} = \frac{1}{9}\)

**Step 4: Solve for \(x\)**

Recall that \(u = 3^x\), so:

1. If \(u = 1\):
   \[
   3^x = 1 \implies x = 0
   \]
   
2. If \(u = \frac{1}{9}\):
   \[
   3^x = \frac{1}{9} = 3^{-2} \implies x = -2
   \]

**Final Solutions:**

\[
x = 0 \quad \text{and} \quad x = -2
\]

Home Question Bank Math Algebra

Question

(9) \( 9^{x+1}-10.3^{x}+1=0 \)

Ask by Clarke Marshall. in South Africa

Jan 26,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The solutions are \( x = 0 \) and \( x = -2 \).

Solution

To solve the equation: \[ 9^{x+1} - 10 \cdot 3^{x} + 1 = 0 \] **Step 1: Express \(9^{x+1}\) in terms of \(3^x\)** Recall that \(9 = 3^2\). Therefore, \[ 9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x + 2} = 3^2 \cdot 3^{2x} = 9 \cdot 3^{2x} \] Substituting back into the original equation: \[ 9 \cdot 3^{2x} - 10 \cdot 3^x + 1 = 0 \] **Step 2: Let \(u = 3^x\)** This substitution simplifies the equation. Let \(u = 3^x\). Then, \(3^{2x} = (3^x)^2 = u^2\). Substituting: \[ 9u^2 - 10u + 1 = 0 \] **Step 3: Solve the quadratic equation for \(u\)** The quadratic equation is: \[ 9u^2 - 10u + 1 = 0 \] Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9\), \(b = -10\), and \(c = 1\): \[ u = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} = \frac{10 \pm \sqrt{100 - 36}}{18} = \frac{10 \pm 8}{18} \] This gives two solutions: 1. \(u = \frac{10 + 8}{18} = \frac{18}{18} = 1\) 2. \(u = \frac{10 - 8}{18} = \frac{2}{18} = \frac{1}{9}\) **Step 4: Solve for \(x\)** Recall that \(u = 3^x\), so: 1. If \(u = 1\): \[ 3^x = 1 \implies x = 0 \] 2. If \(u = \frac{1}{9}\): \[ 3^x = \frac{1}{9} = 3^{-2} \implies x = -2 \] **Final Solutions:** \[ x = 0 \quad \text{and} \quad x = -2 \]

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Beyond the Answer

To solve the equation \( 9^{x+1} - 10 \cdot 3^{x} + 1 = 0 \), let's rewrite \( 9 \) as \( 3^2 \). Thus, \( 9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x + 2} \). Now substituting this in gives us: \[ 3^{2x + 2} - 10 \cdot 3^{x} + 1 = 0 \] Let’s make the substitution \( y = 3^x \). Therefore, \( 3^{2x} = (3^x)^2 = y^2 \), and the equation becomes: \[ y^2 \cdot 3^2 - 10y + 1 = 0 \] Simplifying, we get: \[ 9y^2 - 10y + 1 = 0 \] Now, let’s use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 9, b = -10, c = 1 \): \[ y = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} = \frac{10 \pm \sqrt{100 - 36}}{18} = \frac{10 \pm \sqrt{64}}{18} \] This simplifies to: \[ y = \frac{10 \pm 8}{18} \] Calculating both possibilities: 1. \( y = \frac{18}{18} = 1 \) 2. \( y = \frac{2}{18} = \frac{1}{9} \) Next, we substitute back for \( y \): 1. If \( y = 1 \), then \( 3^x = 1 \) implies \( x = 0 \). 2. If \( y = \frac{1}{9} \), then \( 3^x = \frac{1}{9} \) implies \( x = -2 \). Thus, the solutions to the original equation are: \[ x = 0 \quad \text{and} \quad x = -2 \]

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