Solve for n (81^(3n+2))/(243^(-n))=3^4

$\frac{81^{3 n + 2}}{243^{- n}} = 3^{4}$

Step 1

Move $243^{- n}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$81^{3 n + 2} \cdot 243^{n} = 3^{4}$

Step 2

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

${(3^{4})}^{3 n + 2} \cdot 243^{n} = 3^{4}$

Step 3

Multiply the exponents in ${(3^{4})}^{3 n + 2}$ .

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

$3^{4 (3 n + 2)} \cdot 243^{n} = 3^{4}$

Apply the distributive property.

$3^{4 (3 n) + 4 \cdot 2} \cdot 243^{n} = 3^{4}$

Multiply $3$ by $4$ .

$3^{12 n + 4 \cdot 2} \cdot 243^{n} = 3^{4}$

Multiply $4$ by $2$ .

$3^{12 n + 8} \cdot 243^{n} = 3^{4}$

Step 4

Rewrite $243$ as $3^{5}$ .

$3^{12 n + 8} \cdot {(3^{5})}^{n} = 3^{4}$

Step 5

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

$3^{12 n + 8} \cdot 3^{5 n} = 3^{4}$

Step 6

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

$3^{12 n + 8 + 5 n} = 3^{4}$

Step 7

Add $12 n$ and $5 n$ .

$3^{17 n + 8} = 3^{4}$

Step 8

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

$17 n + 8 = 4$

Step 9

Solve for $n$ .

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

Subtract $8$ from both sides of the equation.

$17 n = 4 - 8$

Subtract $8$ from $4$ .

$17 n = - 4$

Divide each term in $17 n = - 4$ by $17$ and simplify.

Divide each term in $17 n = - 4$ by $17$ .

$\frac{17 n}{17} = \frac{- 4}{17}$

Simplify the left side.

Cancel the common factor of $17$ .

Cancel the common factor.

$\frac{17 n}{17} = \frac{- 4}{17}$

Divide $n$ by $1$ .

$n = \frac{- 4}{17}$

Simplify the right side.

Move the negative in front of the fraction.

$n = - \frac{4}{17}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$n = - \frac{4}{17}$

Decimal Form:

$n = - 0.23529411 \dots$

Solve for n (81^(3n+2))/(243^(-n))=3^4

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