Solve for x 2*4^(x+1)=1/32

$2 \cdot 4^{x + 1} = \frac{1}{32}$

Step 1

Raise $32$ to the power of $1$ .

$2 \cdot 4^{x + 1} = \frac{1}{32^{1}}$

Step 2

Move $32^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$2 \cdot 4^{x + 1} = 32^{- 1}$

Step 3

Rewrite $4$ as $2^{2}$ .

$2 \cdot {(2^{2})}^{x + 1} = 32^{- 1}$

Step 4

Multiply the exponents in ${(2^{2})}^{x + 1}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$2 \cdot 2^{2 (x + 1)} = 32^{- 1}$

Apply the distributive property.

$2 \cdot 2^{2 x + 2 \cdot 1} = 32^{- 1}$

Multiply $2$ by $1$ .

$2 \cdot 2^{2 x + 2} = 32^{- 1}$

Step 5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2^{1 + 2 x + 2} = 32^{- 1}$

Step 6

Add $1$ and $2$ .

$2^{2 x + 3} = 32^{- 1}$

Step 7

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

$2^{2 x + 3} = 2^{- 5}$

Step 8

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

$2 x + 3 = - 5$

Step 9

Solve for $x$ .

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

Subtract $3$ from both sides of the equation.

$2 x = - 5 - 3$

Subtract $3$ from $- 5$ .

$2 x = - 8$

Divide each term in $2 x = - 8$ by $2$ and simplify.

Divide each term in $2 x = - 8$ by $2$ .

$\frac{2 x}{2} = \frac{- 8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 8}{2}$

Divide $x$ by $1$ .

$x = \frac{- 8}{2}$

Simplify the right side.

Divide $- 8$ by $2$ .

$x = - 4$

Solve for x 2*4^(x+1)=1/32

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