Solve for n (1/64)^(3n)*8=2^6

${(\frac{1}{64})}^{3 n} \cdot 8 = 2^{6}$

Apply the product rule to $\frac{1}{64}$ .

$\frac{1^{3 n}}{64^{3 n}} \cdot 8 = 2^{6}$

One to any power is one.

$\frac{1}{64^{3 n}} \cdot 8 = 2^{6}$

Move $64^{3 n}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$64^{- 3 n} \cdot 8 = 2^{6}$

Rewrite $64$ as $2^{6}$ .

${(2^{6})}^{- 3 n} \cdot 8 = 2^{6}$

Multiply the exponents in ${(2^{6})}^{- 3 n}$ .

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

$2^{6 (- 3 n)} \cdot 8 = 2^{6}$

Multiply $- 3$ by $6$ .

$2^{- 18 n} \cdot 8 = 2^{6}$

Rewrite $8$ as $2^{3}$ .

$2^{- 18 n} \cdot 2^{3} = 2^{6}$

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

$2^{- 18 n + 3} = 2^{6}$

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

$- 18 n + 3 = 6$

Solve for $n$ .

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

Subtract $3$ from both sides of the equation.

$- 18 n = 6 - 3$

Subtract $3$ from $6$ .

$- 18 n = 3$

Divide each term in $- 18 n = 3$ by $- 18$ and simplify.

Divide each term in $- 18 n = 3$ by $- 18$ .

$\frac{- 18 n}{- 18} = \frac{3}{- 18}$

Simplify the left side.

Cancel the common factor of $- 18$ .

Cancel the common factor.

$\frac{- 18 n}{- 18} = \frac{3}{- 18}$

Divide $n$ by $1$ .

$n = \frac{3}{- 18}$

Simplify the right side.

Cancel the common factor of $3$ and $- 18$ .

Factor $3$ out of $3$ .

$n = \frac{3 (1)}{- 18}$

Cancel the common factors.

Factor $3$ out of $- 18$ .

$n = \frac{3 \cdot 1}{3 \cdot - 6}$

Cancel the common factor.

$n = \frac{3 \cdot 1}{3 \cdot - 6}$

Rewrite the expression.

$n = \frac{1}{- 6}$

Move the negative in front of the fraction.

$n = - \frac{1}{6}$

The result can be shown in multiple forms.

Exact Form:

$n = - \frac{1}{6}$

Decimal Form:

$n = - 0.1\overline{6}$

Solve for n (1/64)^(3n)*8=2^6

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