Precalculus Examples

$7^{2 x} + 7^{x} - 12 = 0$

Step 1

Rewrite $7^{2 x}$ as ${(7^{x})}^{2}$ .

${(7^{x})}^{2} + 7^{x} - 12 = 0$

Step 2

Let $u = 7^{x}$ . Substitute $u$ for all occurrences of $7^{x}$ .

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

Step 3

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

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Step 3.1

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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Step 3.2

Write the factored form using these integers.

$(u - 3) (u + 4) = 0$

Step 4

Replace all occurrences of $u$ with $7^{x}$ .

$(7^{x} - 3) (7^{x} + 4) = 0$

Step 5

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

$7^{x} - 3 = 0$

$7^{x} + 4 = 0$

Step 6

Set $7^{x} - 3$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $7^{x} - 3$ equal to $0$ .

$7^{x} - 3 = 0$

Step 6.2

Solve $7^{x} - 3 = 0$ for $x$ .

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Step 6.2.1

Add $3$ to both sides of the equation.

$7^{x} = 3$

Step 6.2.2

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

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

Step 6.2.3

Expand $\ln (7^{x})$ by moving $x$ outside the logarithm.

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

Step 6.2.4

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

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Step 6.2.4.1

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

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

Step 6.2.4.2

Simplify the left side.

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Step 6.2.4.2.1

Cancel the common factor of $\ln (7)$ .

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Step 6.2.4.2.1.1

Cancel the common factor.

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

Step 6.2.4.2.1.2

Divide $x$ by $1$ .

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

Step 7

Set $7^{x} + 4$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $7^{x} + 4$ equal to $0$ .

$7^{x} + 4 = 0$

Step 7.2

Solve $7^{x} + 4 = 0$ for $x$ .

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Step 7.2.1

Subtract $4$ from both sides of the equation.

$7^{x} = - 4$

Step 7.2.2

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

$\ln (7^{x}) = \ln (- 4)$

Step 7.2.3

The equation cannot be solved because $\ln (- 4)$ is undefined.

Undefined

Step 7.2.4

There is no solution for $7^{x} = - 4$

No solution

Step 8

The final solution is all the values that make $(7^{x} - 3) (7^{x} + 4) = 0$ true.

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