Calculus Examples

$2^{2 x} - 32 \cdot 2^{x} + 256 = 0$

Step 1

Rewrite $2^{2 x}$ as exponentiation.

${(2^{x})}^{2} - 32 \cdot 2^{x} + 256 = 0$

Step 2

Substitute $u$ for $2^{x}$ .

$u^{2} - 32 u + 256 = 0$

Step 3

Solve for $u$ .

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Step 3.1

Factor using the perfect square rule.

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Step 3.1.1

Rewrite $256$ as $16^{2}$ .

$u^{2} - 32 u + 16^{2} = 0$

Step 3.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$32 u = 2 \cdot u \cdot 16$

Step 3.1.3

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 16 + 16^{2} = 0$

Step 3.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 16$ .

${(u - 16)}^{2} = 0$

Step 3.2

Set the $u - 16$ equal to $0$ .

$u - 16 = 0$

Step 3.3

Add $16$ to both sides of the equation.

$u = 16$

Step 4

Substitute $16$ for $u$ in $u = 2^{x}$ .

$16 = 2^{x}$

Step 5

Solve $16 = 2^{x}$ .

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Step 5.1

Rewrite the equation as $2^{x} = 16$ .

$2^{x} = 16$

Step 5.2

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

$2^{x} = 2^{4}$

Step 5.3

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

$x = 4$