Precalculus Examples

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

Step 1

Factor the left side of the equation.

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Step 1.1

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

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

Step 1.2

Let $u = 2^{x}$ . Substitute $u$ for all occurrences of $2^{x}$ .

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

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 $- 1$ .

$- 3, 2$

Step 1.3.2

Write the factored form using these integers.

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

Step 1.4

Replace all occurrences of $u$ with $2^{x}$ .

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

Step 2

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

$2^{x} - 3 = 0$

$2^{x} + 2 = 0$

Step 3

Set $2^{x} - 3$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $2^{x} - 3$ equal to $0$ .

$2^{x} - 3 = 0$

Step 3.2

Solve $2^{x} - 3 = 0$ for $x$ .

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Step 3.2.1

Add $3$ to both sides of the equation.

$2^{x} = 3$

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.2.4

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

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Step 3.2.4.1

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

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

Step 3.2.4.2

Simplify the left side.

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Step 3.2.4.2.1

Cancel the common factor of $\ln (2)$ .

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Step 3.2.4.2.1.1

Cancel the common factor.

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

Step 3.2.4.2.1.2

Divide $x$ by $1$ .

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

Step 4

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

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Step 4.1

Set $2^{x} + 2$ equal to $0$ .

$2^{x} + 2 = 0$

Step 4.2

Solve $2^{x} + 2 = 0$ for $x$ .

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Step 4.2.1

Subtract $2$ from both sides of the equation.

$2^{x} = - 2$

Step 4.2.2

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

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

Step 4.2.3

The equation cannot be solved because $\ln (- 2)$ is undefined.

Undefined

Step 4.2.4

There is no solution for $2^{x} = - 2$

No solution

Step 5

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

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 1.58496250 \dots$