Solve for x 16^(x+1)=8^(2x+1)

$16^{x + 1} = 8^{2 x + 1}$

Step 1

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

$2^{4 (x + 1)} = 2^{3 (2 x + 1)}$

Step 2

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

$4 (x + 1) = 3 (2 x + 1)$

Step 3

Solve for $x$ .

Simplify $4 (x + 1)$ .

Rewrite.

$0 + 0 + 4 (x + 1) = 3 (2 x + 1)$

Simplify by adding zeros.

$4 (x + 1) = 3 (2 x + 1)$

Apply the distributive property.

$4 x + 4 \cdot 1 = 3 (2 x + 1)$

Multiply $4$ by $1$ .

$4 x + 4 = 3 (2 x + 1)$

Simplify $3 (2 x + 1)$ .

Apply the distributive property.

$4 x + 4 = 3 (2 x) + 3 \cdot 1$

Multiply.

Multiply $2$ by $3$ .

$4 x + 4 = 6 x + 3 \cdot 1$

Multiply $3$ by $1$ .

$4 x + 4 = 6 x + 3$

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

Subtract $6 x$ from both sides of the equation.

$4 x + 4 - 6 x = 3$

Subtract $6 x$ from $4 x$ .

$- 2 x + 4 = 3$

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

Subtract $4$ from both sides of the equation.

$- 2 x = 3 - 4$

Subtract $4$ from $3$ .

$- 2 x = - 1$

Divide each term in $- 2 x = - 1$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 1$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 1}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 1}{- 2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{1}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{2}$

Decimal Form:

$x = 0.5$

Solve for x 16^(x+1)=8^(2x+1)

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