Solve the Rational Equation for x 4^x+2^(x+1)=8

$4^{x} + 2^{x + 1} = 8$

Step 1

Subtract $8$ from both sides of the equation.

$4^{x} + 2^{x + 1} - 8 = 0$

Step 2

Factor the left side of the equation.

Rewrite $2^{x + 1}$ as $2^{x} \cdot 2^{1}$ .

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

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

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

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

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

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

Evaluate the exponent.

$u^{2} + u \cdot 2 - 8 = 0$

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

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

Factor $u^{2} + 2 u - 8$ 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 $- 8$ and whose sum is $2$ .

$- 2, 4$

Write the factored form using these integers.

$(u - 2) (u + 4) = 0$

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

$(2^{x} - 2) (2^{x} + 4) = 0$

Step 3

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

$2^{x} - 2 = 0$

$2^{x} + 4 = 0$

Step 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 $x$ .

Add $2$ to both sides of the equation.

$2^{x} = 2$

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

$2^{x} = 2^{1}$

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

$x = 1$

Step 5

Set $2^{x} + 4$ equal to $0$ and solve for $x$ .

Set $2^{x} + 4$ equal to $0$ .

$2^{x} + 4 = 0$

Solve $2^{x} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$2^{x} = - 4$

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

$\ln (2^{x}) = \ln (- 4)$

The equation cannot be solved because $\ln (- 4)$ is undefined.

Undefined

There is no solution for $2^{x} = - 4$

No solution

Step 6

The final solution is all the values that make $(2^{x} - 2) (2^{x} + 4) = 0$ true.

$x = 1$

Solve the Rational Equation for x 4^x+2^(x+1)=8

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