Solve for x 4^x-14*2^x-32=0

$4^{x} - 14 \cdot 2^{x} - 32 = 0$

Factor the left side of the equation.

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

${(2^{2})}^{x} - 14 \cdot 2^{x} - 32 = 0$

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

${(2^{x})}^{2} - 14 \cdot 2^{x} - 32 = 0$

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

$u^{2} - 14 u - 32 = 0$

Factor $u^{2} - 14 u - 32$ using the AC method.

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 $- 32$ and whose sum is $- 14$ .

$- 16, 2$

Write the factored form using these integers.

$(u - 16) (u + 2) = 0$

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

$(2^{x} - 16) (2^{x} + 2) = 0$

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

$2^{x} - 16 = 0$

$2^{x} + 2 = 0$

Set $2^{x} - 16$ equal to $0$ and solve for $x$ .

Set $2^{x} - 16$ equal to $0$ .

$2^{x} - 16 = 0$

Solve $2^{x} - 16 = 0$ for $x$ .

Add $16$ to both sides of the equation.

$2^{x} = 16$

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

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

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

$x = 4$

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

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

$2^{x} + 2 = 0$

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

Subtract $2$ from both sides of the equation.

$2^{x} = - 2$

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

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

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

Undefined

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

No solution

The final solution is all the values that make $(2^{x} - 16) (2^{x} + 2) = 0$ true.

$x = 4$

Solve for x 4^x-14*2^x-32=0

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