Precalculus Examples

$3^{2 x} + 35 = 12 (3^{x})$

Step 1

Subtract $12 \cdot 3^{x}$ from both sides of the equation.

$3^{2 x} + 35 - 12 \cdot 3^{x} = 0$

Step 2

Rewrite $3^{2 x}$ as exponentiation.

${(3^{x})}^{2} + 35 - 12 \cdot 3^{x} = 0$

Step 3

Substitute $u$ for $3^{x}$ .

$u^{2} + 35 - 12 u = 0$

Step 4

Move $35$ .

$u^{2} - 12 u + 35 = 0$

Step 5

Solve for $u$ .

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Step 5.1

Factor $u^{2} - 12 u + 35$ using the AC method.

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Step 5.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $35$ and whose sum is $- 12$ .

$- 7, - 5$

Step 5.1.2

Write the factored form using these integers.

$(u - 7) (u - 5) = 0$

Step 5.2

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

$u - 7 = 0$

$u - 5 = 0$

Step 5.3

Set $u - 7$ equal to $0$ and solve for $u$ .

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Step 5.3.1

Set $u - 7$ equal to $0$ .

$u - 7 = 0$

Step 5.3.2

Add $7$ to both sides of the equation.

$u = 7$

Step 5.4

Set $u - 5$ equal to $0$ and solve for $u$ .

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Step 5.4.1

Set $u - 5$ equal to $0$ .

$u - 5 = 0$

Step 5.4.2

Add $5$ to both sides of the equation.

$u = 5$

Step 5.5

The final solution is all the values that make $(u - 7) (u - 5) = 0$ true.

$u = 7, 5$

Step 6

Substitute $7$ for $u$ in $u = 3^{x}$ .

$7 = 3^{x}$

Step 7

Solve $7 = 3^{x}$ .

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Step 7.1

Rewrite the equation as $3^{x} = 7$ .

$3^{x} = 7$

Step 7.2

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

$\ln (3^{x}) = \ln (7)$

Step 7.3

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

$x \ln (3) = \ln (7)$

Step 7.4

Divide each term in $x \ln (3) = \ln (7)$ by $\ln (3)$ and simplify.

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Step 7.4.1

Divide each term in $x \ln (3) = \ln (7)$ by $\ln (3)$ .

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (7)}{\ln (3)}$

Step 7.4.2

Simplify the left side.

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Step 7.4.2.1

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

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Step 7.4.2.1.1

Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (7)}{\ln (3)}$

Step 7.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (7)}{\ln (3)}$

Step 8

Substitute $5$ for $u$ in $u = 3^{x}$ .

$5 = 3^{x}$

Step 9

Solve $5 = 3^{x}$ .

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Step 9.1

Rewrite the equation as $3^{x} = 5$ .

$3^{x} = 5$

Step 9.2

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

$\ln (3^{x}) = \ln (5)$

Step 9.3

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

$x \ln (3) = \ln (5)$

Step 9.4

Divide each term in $x \ln (3) = \ln (5)$ by $\ln (3)$ and simplify.

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Step 9.4.1

Divide each term in $x \ln (3) = \ln (5)$ by $\ln (3)$ .

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (5)}{\ln (3)}$

Step 9.4.2

Simplify the left side.

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Step 9.4.2.1

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

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Step 9.4.2.1.1

Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (5)}{\ln (3)}$

Step 9.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (5)}{\ln (3)}$

Step 10

List the solutions that makes the equation true.

$x = \frac{\ln (7)}{\ln (3)}, \frac{\ln (5)}{\ln (3)}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (7)}{\ln (3)}, \frac{\ln (5)}{\ln (3)}$

Decimal Form:

$x = 1.77124374 \dots, 1.46497352 \dots$