Solve for x (1/2)^(x+1)=8^x

${(\frac{1}{2})}^{x + 1} = 8^{x}$

Step 1

Apply the product rule to $\frac{1}{2}$ .

$\frac{1^{x + 1}}{2^{x + 1}} = 8^{x}$

Step 2

One to any power is one.

$\frac{1}{2^{x + 1}} = 8^{x}$

Step 3

Move $2^{x + 1}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$2^{- (x + 1)} = 8^{x}$

Step 4

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

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

Step 5

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

$- (x + 1) = 3 x$

Step 6

Solve for $x$ .

Simplify $- (x + 1)$ .

Start simplifying.

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

Simplify by adding zeros.

$- (x + 1) = 3 x$

Apply the distributive property.

$- x - 1 \cdot 1 = 3 x$

Multiply $- 1$ by $1$ .

$- x - 1 = 3 x$

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

Subtract $3 x$ from both sides of the equation.

$- x - 1 - 3 x = 0$

Subtract $3 x$ from $- x$ .

$- 4 x - 1 = 0$

Add $1$ to both sides of the equation.

$- 4 x = 1$

Divide each term in $- 4 x = 1$ by $- 4$ and simplify.

Divide each term in $- 4 x = 1$ by $- 4$ .

$\frac{- 4 x}{- 4} = \frac{1}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x}{- 4} = \frac{1}{- 4}$

Divide $x$ by $1$ .

$x = \frac{1}{- 4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{4}$

Decimal Form:

$x = - 0.25$

Solve for x (1/2)^(x+1)=8^x

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