Precalculus Examples

${(4^{x})}^{2} - 65 \cdot 4^{x} + 64 = 0$

Step 1

Simplify the left side.

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Step 1.1

Multiply the exponents in ${(4^{x})}^{2}$ .

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Step 1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4^{x \cdot 2} - 65 \cdot 4^{x} + 64 = 0$

Step 1.1.2

Move $2$ to the left of $x$ .

$4^{2 x} - 65 \cdot 4^{x} + 64 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

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

${(4^{x})}^{2} - 65 \cdot 4^{x} + 64 = 0$

Step 2.2

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

$u^{2} - 65 u + 64 = 0$

Step 2.3

Factor $u^{2} - 65 u + 64$ using the AC method.

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Step 2.3.1

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 $64$ and whose sum is $- 65$ .

$- 64, - 1$

Step 2.3.2

Write the factored form using these integers.

$(u - 64) (u - 1) = 0$

Step 2.4

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

$(4^{x} - 64) (4^{x} - 1) = 0$

Step 3

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

$4^{x} - 64 = 0$

$4^{x} - 1 = 0$

Step 4

Set $4^{x} - 64$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $4^{x} - 64$ equal to $0$ .

$4^{x} - 64 = 0$

Step 4.2

Solve $4^{x} - 64 = 0$ for $x$ .

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Step 4.2.1

Add $64$ to both sides of the equation.

$4^{x} = 64$

Step 4.2.2

Create equivalent expressions in the equation that all have equal bases.

$4^{x} = 4^{3}$

Step 4.2.3

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$x = 3$

Step 5

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

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Step 5.1

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

$4^{x} - 1 = 0$

Step 5.2

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

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Step 5.2.1

Add $1$ to both sides of the equation.

$4^{x} = 1$

Step 5.2.2

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

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

Step 5.2.3

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

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

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

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

$x \ln (4) = 0$

Step 5.2.5

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

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Step 5.2.5.1

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

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

Step 5.2.5.2

Simplify the left side.

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Step 5.2.5.2.1

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

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Step 5.2.5.2.1.1

Cancel the common factor.

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

Step 5.2.5.2.1.2

Divide $x$ by $1$ .

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

Step 5.2.5.3

Simplify the right side.

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Step 5.2.5.3.1

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

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

Step 5.2.5.3.2

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

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

Step 5.2.5.3.3

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

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Step 5.2.5.3.3.1

Factor $2$ out of $0$ .

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

Step 5.2.5.3.3.2

Cancel the common factors.

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Step 5.2.5.3.3.2.1

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

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

Step 5.2.5.3.3.2.2

Cancel the common factor.

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

Step 5.2.5.3.3.2.3

Rewrite the expression.

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

Step 5.2.5.3.4

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

$x = 0$

Step 6

The final solution is all the values that make $(4^{x} - 64) (4^{x} - 1) = 0$ true.

$x = 3, 0$