Precalculus Examples

$16^{x} - 8 \cdot 4^{x} + 7 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Rewrite $16^{x}$ as ${(4^{2})}^{x}$ .

${(4^{2})}^{x} - 8 \cdot 4^{x} + 7 = 0$

Step 1.2

Rewrite ${(4^{2})}^{x}$ as ${(4^{x})}^{2}$ .

${(4^{x})}^{2} - 8 \cdot 4^{x} + 7 = 0$

Step 1.3

Let $u = 4^{x}$ . Substitute $u$ for all occurrences of $4^{x}$ .

$u^{2} - 8 u + 7 = 0$

Step 1.4

Factor $u^{2} - 8 u + 7$ using the AC method.

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Step 1.4.1

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 $7$ and whose sum is $- 8$ .

$- 7, - 1$

Step 1.4.2

Write the factored form using these integers.

$(u - 7) (u - 1) = 0$

Step 1.5

Replace all occurrences of $u$ with $4^{x}$ .

$(4^{x} - 7) (4^{x} - 1) = 0$

Step 2

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

$4^{x} - 7 = 0$

$4^{x} - 1 = 0$

Step 3

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

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Step 3.1

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

$4^{x} - 7 = 0$

Step 3.2

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

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Step 3.2.1

Add $7$ to both sides of the equation.

$4^{x} = 7$

Step 3.2.2

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

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

Step 3.2.3

Expand $\ln (4^{x})$ by moving $x$ outside the logarithm.

$x \ln (4) = \ln (7)$

Step 3.2.4

Divide each term in $x \ln (4) = \ln (7)$ by $\ln (4)$ and simplify.

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Step 3.2.4.1

Divide each term in $x \ln (4) = \ln (7)$ by $\ln (4)$ .

$\frac{x \ln (4)}{\ln (4)} = \frac{\ln (7)}{\ln (4)}$

Step 3.2.4.2

Simplify the left side.

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Step 3.2.4.2.1

Cancel the common factor of $\ln (4)$ .

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Step 3.2.4.2.1.1

Cancel the common factor.

$\frac{x \ln (4)}{\ln (4)} = \frac{\ln (7)}{\ln (4)}$

Step 3.2.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (7)}{\ln (4)}$

Step 4

Set $4^{x} - 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $4^{x} - 1$ equal to $0$ .

$4^{x} - 1 = 0$

Step 4.2

Solve $4^{x} - 1 = 0$ for $x$ .

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Step 4.2.1

Add $1$ to both sides of the equation.

$4^{x} = 1$

Step 4.2.2

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

$\ln (4^{x}) = \ln (1)$

Step 4.2.3

Expand $\ln (4^{x})$ by moving $x$ outside the logarithm.

$x \ln (4) = \ln (1)$

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

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

$x \ln (4) = 0$

Step 4.2.5

Divide each term in $x \ln (4) = 0$ by $\ln (4)$ and simplify.

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Step 4.2.5.1

Divide each term in $x \ln (4) = 0$ by $\ln (4)$ .

$\frac{x \ln (4)}{\ln (4)} = \frac{0}{\ln (4)}$

Step 4.2.5.2

Simplify the left side.

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Step 4.2.5.2.1

Cancel the common factor of $\ln (4)$ .

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Step 4.2.5.2.1.1

Cancel the common factor.

$\frac{x \ln (4)}{\ln (4)} = \frac{0}{\ln (4)}$

Step 4.2.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{\ln (4)}$

Step 4.2.5.3

Simplify the right side.

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Step 4.2.5.3.1

Rewrite $\ln (4)$ as $\ln (2^{2})$ .

$x = \frac{0}{\ln (2^{2})}$

Step 4.2.5.3.2

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

$x = \frac{0}{2 \ln (2)}$

Step 4.2.5.3.3

Cancel the common factor of $0$ and $2$ .

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Step 4.2.5.3.3.1

Factor $2$ out of $0$ .

$x = \frac{2 (0)}{2 \ln (2)}$

Step 4.2.5.3.3.2

Cancel the common factors.

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Step 4.2.5.3.3.2.1

Factor $2$ out of $2 \ln (2)$ .

$x = \frac{2 (0)}{2 (\ln (2))}$

Step 4.2.5.3.3.2.2

Cancel the common factor.

$x = \frac{2 \cdot 0}{2 \ln (2)}$

Step 4.2.5.3.3.2.3

Rewrite the expression.

$x = \frac{0}{\ln (2)}$

Step 4.2.5.3.4

Divide $0$ by $\ln (2)$ .

$x = 0$

Step 5

The final solution is all the values that make $(4^{x} - 7) (4^{x} - 1) = 0$ true.

$x = \frac{\ln (7)}{\ln (4)}, 0$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (7)}{\ln (4)}, 0$

Decimal Form:

$x = 1.40367746 \dots, 0$