Solve for a (1/9)^(a+1)=81^(a+1)*27^(2-a)

${(\frac{1}{9})}^{a + 1} = 81^{a + 1} \cdot 27^{2 - a}$

Step 1

Apply the product rule to $\frac{1}{9}$ .

$\frac{1^{a + 1}}{9^{a + 1}} = 81^{a + 1} \cdot 27^{2 - a}$

Step 2

One to any power is one.

$\frac{1}{9^{a + 1}} = 81^{a + 1} \cdot 27^{2 - a}$

Step 3

Move $9^{a + 1}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$9^{- (a + 1)} = 81^{a + 1} \cdot 27^{2 - a}$

Step 4

Rewrite $81$ as $3^{4}$ .

$9^{- (a + 1)} = {(3^{4})}^{a + 1} \cdot 27^{2 - a}$

Step 5

Multiply the exponents in ${(3^{4})}^{a + 1}$ .

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

$9^{- (a + 1)} = 3^{4 (a + 1)} \cdot 27^{2 - a}$

Apply the distributive property.

$9^{- (a + 1)} = 3^{4 a + 4 \cdot 1} \cdot 27^{2 - a}$

Multiply $4$ by $1$ .

$9^{- (a + 1)} = 3^{4 a + 4} \cdot 27^{2 - a}$

Step 6

Rewrite $27$ as $3^{3}$ .

$9^{- (a + 1)} = 3^{4 a + 4} \cdot {(3^{3})}^{2 - a}$

Step 7

Multiply the exponents in ${(3^{3})}^{2 - a}$ .

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

$9^{- (a + 1)} = 3^{4 a + 4} \cdot 3^{3 (2 - a)}$

Apply the distributive property.

$9^{- (a + 1)} = 3^{4 a + 4} \cdot 3^{3 \cdot 2 + 3 (- a)}$

Multiply $3$ by $2$ .

$9^{- (a + 1)} = 3^{4 a + 4} \cdot 3^{6 + 3 (- a)}$

Multiply $- 1$ by $3$ .

$9^{- (a + 1)} = 3^{4 a + 4} \cdot 3^{6 - 3 a}$

Step 8

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

$9^{- (a + 1)} = 3^{4 a + 4 + 6 - 3 a}$

Step 9

Subtract $3 a$ from $4 a$ .

$9^{- (a + 1)} = 3^{a + 4 + 6}$

Step 10

Add $4$ and $6$ .

$9^{- (a + 1)} = 3^{a + 10}$

Step 11

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

$3^{2 (- (a + 1))} = 3^{a + 10}$

Step 12

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

$2 (- (a + 1)) = a + 10$

Step 13

Solve for $a$ .

Simplify $2 (- (a + 1))$ .

Rewrite.

$0 + 0 + 2 (- (a + 1)) = a + 10$

Simplify by adding zeros.

$2 (- (a + 1)) = a + 10$

Apply the distributive property.

$2 (- a - 1 \cdot 1) = a + 10$

Multiply $- 1$ by $1$ .

$2 (- a - 1) = a + 10$

Apply the distributive property.

$2 (- a) + 2 \cdot - 1 = a + 10$

Multiply.

Multiply $- 1$ by $2$ .

$- 2 a + 2 \cdot - 1 = a + 10$

Multiply $2$ by $- 1$ .

$- 2 a - 2 = a + 10$

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

Subtract $a$ from both sides of the equation.

$- 2 a - 2 - a = 10$

Subtract $a$ from $- 2 a$ .

$- 3 a - 2 = 10$

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

Add $2$ to both sides of the equation.

$- 3 a = 10 + 2$

Add $10$ and $2$ .

$- 3 a = 12$

Divide each term in $- 3 a = 12$ by $- 3$ and simplify.

Divide each term in $- 3 a = 12$ by $- 3$ .

$\frac{- 3 a}{- 3} = \frac{12}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 a}{- 3} = \frac{12}{- 3}$

Divide $a$ by $1$ .

$a = \frac{12}{- 3}$

Simplify the right side.

Divide $12$ by $- 3$ .

$a = - 4$

Solve for a (1/9)^(a+1)=81^(a+1)*27^(2-a)

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