Trigonometry Examples

$4^{2 x} - 63 \cdot 4^{x} - 64 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

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

${(4^{x})}^{2} - 63 \cdot 4^{x} - 64 = 0$

Step 1.2

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

$u^{2} - 63 u - 64 = 0$

Step 1.3

Factor $u^{2} - 63 u - 64$ 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 $- 64$ and whose sum is $- 63$ .

$- 64, 1$

Step 1.3.2

Write the factored form using these integers.

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

Step 1.4

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

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

Step 2

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

$4^{x} - 64 = 0$

$4^{x} + 1 = 0$

Step 3

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

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Step 3.1

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

$4^{x} - 64 = 0$

Step 3.2

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

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Step 3.2.1

Add $64$ to both sides of the equation.

$4^{x} = 64$

Step 3.2.2

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

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

Step 3.2.3

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

$x = 3$

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

Subtract $1$ from 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

The equation cannot be solved because $\ln (- 1)$ is undefined.

Undefined

Step 4.2.4

There is no solution for $4^{x} = - 1$

No solution

Step 5

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

$x = 3$