Solve for x 8^(1-x)=4^(x+2)

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

Step 1

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

$2^{3 (1 - x)} = 2^{2 (x + 2)}$

Step 2

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

$3 (1 - x) = 2 (x + 2)$

Step 3

Solve for $x$ .

Simplify $3 (1 - x)$ .

Rewrite.

$0 + 0 + 3 (1 - x) = 2 (x + 2)$

Simplify by adding zeros.

$3 (1 - x) = 2 (x + 2)$

Apply the distributive property.

$3 \cdot 1 + 3 (- x) = 2 (x + 2)$

Multiply.

Multiply $3$ by $1$ .

$3 + 3 (- x) = 2 (x + 2)$

Multiply $- 1$ by $3$ .

$3 - 3 x = 2 (x + 2)$

Simplify $2 (x + 2)$ .

Apply the distributive property.

$3 - 3 x = 2 x + 2 \cdot 2$

Multiply $2$ by $2$ .

$3 - 3 x = 2 x + 4$

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$3 - 3 x - 2 x = 4$

Subtract $2 x$ from $- 3 x$ .

$3 - 5 x = 4$

Move all terms not containing $x$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$- 5 x = 4 - 3$

Subtract $3$ from $4$ .

$- 5 x = 1$

Divide each term in $- 5 x = 1$ by $- 5$ and simplify.

Divide each term in $- 5 x = 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{1}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{1}{- 5}$

Divide $x$ by $1$ .

$x = \frac{1}{- 5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{5}$

Decimal Form:

$x = - 0.2$

Solve for x 8^(1-x)=4^(x+2)

Related Questions

Support LCMGCF.com by disabling your adblocker.