Precalculus Examples

$9 (9^{x}) - 10 \cdot 3^{x} + 1 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Rewrite $9^{x}$ as ${(3^{2})}^{x}$ .

$9 \cdot {(3^{2})}^{x} - 10 \cdot 3^{x} + 1 = 0$

Step 1.2

Rewrite ${(3^{2})}^{x}$ as ${(3^{x})}^{2}$ .

$9 \cdot {(3^{x})}^{2} - 10 \cdot 3^{x} + 1 = 0$

Step 1.3

Let $u = 3^{x}$ . Substitute $u$ for all occurrences of $3^{x}$ .

$9 u^{2} - 10 u + 1 = 0$

Step 1.4

Factor by grouping.

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Step 1.4.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 9 \cdot 1 = 9$ and whose sum is $b = - 10$ .

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Step 1.4.1.1

Factor $- 10$ out of $- 10 u$ .

$9 u^{2} - 10 u + 1 = 0$

Step 1.4.1.2

Rewrite $- 10$ as $- 1$ plus $- 9$

$9 u^{2} + (- 1 - 9) u + 1 = 0$

Step 1.4.1.3

Apply the distributive property.

$9 u^{2} - 1 u - 9 u + 1 = 0$

Step 1.4.2

Factor out the greatest common factor from each group.

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Step 1.4.2.1

Group the first two terms and the last two terms.

$(9 u^{2} - 1 u) - 9 u + 1 = 0$

Step 1.4.2.2

Factor out the greatest common factor (GCF) from each group.

$u (9 u - 1) - (9 u - 1) = 0$

Step 1.4.3

Factor the polynomial by factoring out the greatest common factor, $9 u - 1$ .

$(9 u - 1) (u - 1) = 0$

Step 1.5

Replace all occurrences of $u$ with $3^{x}$ .

$(9 \cdot 3^{x} - 1) (3^{x} - 1) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$9 \cdot 3^{x} - 1 = 0$

$3^{x} - 1 = 0$

Step 3

Set $9 \cdot 3^{x} - 1$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $9 \cdot 3^{x} - 1$ equal to $0$ .

$9 \cdot 3^{x} - 1 = 0$

Step 3.2

Solve $9 \cdot 3^{x} - 1 = 0$ for $x$ .

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Step 3.2.1

Multiply $9 \cdot 3^{x}$ .

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Step 3.2.1.1

Rewrite $9$ as $3^{2}$ .

$3^{2} \cdot 3^{x} - 1 = 0$

Step 3.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3^{2 + x} - 1 = 0$

Step 3.2.2

Add $1$ to both sides of the equation.

$3^{2 + x} = 1$

Step 3.2.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (3^{2 + x}) = \ln (1)$

Step 3.2.4

Expand $\ln (3^{2 + x})$ by moving $2 + x$ outside the logarithm.

$(2 + x) \ln (3) = \ln (1)$

Step 3.2.5

Simplify the left side.

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Step 3.2.5.1

Apply the distributive property.

$2 \ln (3) + x \ln (3) = \ln (1)$

Step 3.2.6

Simplify the right side.

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Step 3.2.6.1

The natural logarithm of $1$ is $0$ .

$2 \ln (3) + x \ln (3) = 0$

Step 3.2.7

Reorder $2 \ln (3)$ and $x \ln (3)$ .

$x \ln (3) + 2 \ln (3) = 0$

Step 3.2.8

Subtract $2 \ln (3)$ from both sides of the equation.

$x \ln (3) = - 2 \ln (3)$

Step 3.2.9

Divide each term in $x \ln (3) = - 2 \ln (3)$ by $\ln (3)$ and simplify.

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Step 3.2.9.1

Divide each term in $x \ln (3) = - 2 \ln (3)$ by $\ln (3)$ .

$\frac{x \ln (3)}{\ln (3)} = \frac{- 2 \ln (3)}{\ln (3)}$

Step 3.2.9.2

Simplify the left side.

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Step 3.2.9.2.1

Cancel the common factor of $\ln (3)$ .

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Step 3.2.9.2.1.1

Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{- 2 \ln (3)}{\ln (3)}$

Step 3.2.9.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 2 \ln (3)}{\ln (3)}$

Step 3.2.9.3

Simplify the right side.

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Step 3.2.9.3.1

Cancel the common factor of $\ln (3)$ .

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Step 3.2.9.3.1.1

Cancel the common factor.

$x = \frac{- 2 \ln (3)}{\ln (3)}$

Step 3.2.9.3.1.2

Divide $- 2$ by $1$ .

$x = - 2$

Step 4

Set $3^{x} - 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $3^{x} - 1$ equal to $0$ .

$3^{x} - 1 = 0$

Step 4.2

Solve $3^{x} - 1 = 0$ for $x$ .

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Step 4.2.1

Add $1$ to both sides of the equation.

$3^{x} = 1$

Step 4.2.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (3^{x}) = \ln (1)$

Step 4.2.3

Expand $\ln (3^{x})$ by moving $x$ outside the logarithm.

$x \ln (3) = \ln (1)$

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

The natural logarithm of $1$ is $0$ .

$x \ln (3) = 0$

Step 4.2.5

Divide each term in $x \ln (3) = 0$ by $\ln (3)$ and simplify.

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Step 4.2.5.1

Divide each term in $x \ln (3) = 0$ by $\ln (3)$ .

$\frac{x \ln (3)}{\ln (3)} = \frac{0}{\ln (3)}$

Step 4.2.5.2

Simplify the left side.

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Step 4.2.5.2.1

Cancel the common factor of $\ln (3)$ .

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Step 4.2.5.2.1.1

Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{0}{\ln (3)}$

Step 4.2.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{\ln (3)}$

Step 4.2.5.3

Simplify the right side.

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Step 4.2.5.3.1

Divide $0$ by $\ln (3)$ .

$x = 0$

Step 5

The final solution is all the values that make $(9 \cdot 3^{x} - 1) (3^{x} - 1) = 0$ true.

$x = - 2, 0$