Precalculus Examples

$7^{2 x} + 7^{x} - 6 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

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

${(7^{x})}^{2} + 7^{x} - 6 = 0$

Step 1.2

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

$u^{2} + u - 6 = 0$

Step 1.3

Factor $u^{2} + u - 6$ using the AC method.

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Step 1.3.1

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Step 1.3.2

Write the factored form using these integers.

$(u - 2) (u + 3) = 0$

Step 1.4

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

$(7^{x} - 2) (7^{x} + 3) = 0$

Step 2

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

$7^{x} - 2 = 0$

$7^{x} + 3 = 0$

Step 3

Set $7^{x} - 2$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $7^{x} - 2$ equal to $0$ .

$7^{x} - 2 = 0$

Step 3.2

Solve $7^{x} - 2 = 0$ for $x$ .

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Step 3.2.1

Add $2$ to both sides of the equation.

$7^{x} = 2$

Step 3.2.2

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

$\ln (7^{x}) = \ln (2)$

Step 3.2.3

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

$x \ln (7) = \ln (2)$

Step 3.2.4

Divide each term in $x \ln (7) = \ln (2)$ by $\ln (7)$ and simplify.

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Step 3.2.4.1

Divide each term in $x \ln (7) = \ln (2)$ by $\ln (7)$ .

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (2)}{\ln (7)}$

Step 3.2.4.2

Simplify the left side.

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Step 3.2.4.2.1

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

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Step 3.2.4.2.1.1

Cancel the common factor.

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (2)}{\ln (7)}$

Step 3.2.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (2)}{\ln (7)}$

Step 4

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

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Step 4.1

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

$7^{x} + 3 = 0$

Step 4.2

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

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Step 4.2.1

Subtract $3$ from both sides of the equation.

$7^{x} = - 3$

Step 4.2.2

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

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

Step 4.2.3

The equation cannot be solved because $\ln (- 3)$ is undefined.

Undefined

Step 4.2.4

There is no solution for $7^{x} = - 3$

No solution

Step 5

The final solution is all the values that make $(7^{x} - 2) (7^{x} + 3) = 0$ true.

$x = \frac{\ln (2)}{\ln (7)}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (2)}{\ln (7)}$

Decimal Form:

$x = 0.35620718 \dots$