Solve for d (1/3)^(d-5)=81

${(\frac{1}{3})}^{d - 5} = 81$

Apply the product rule to $\frac{1}{3}$ .

$\frac{1^{d - 5}}{3^{d - 5}} = 81$

One to any power is one.

$\frac{1}{3^{d - 5}} = 81$

Move $3^{d - 5}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$3^{- (d - 5)} = 81$

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

$3^{- (d - 5)} = 3^{4}$

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

$- (d - 5) = 4$

Solve for $d$ .

Simplify $- (d - 5)$ .

Apply the distributive property.

$- d - - 5 = 4$

Multiply $- 1$ by $- 5$ .

$- d + 5 = 4$

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

Subtract $5$ from both sides of the equation.

$- d = 4 - 5$

Subtract $5$ from $4$ .

$- d = - 1$

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

Divide each term in $- d = - 1$ by $- 1$ .

$\frac{- d}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{d}{1} = \frac{- 1}{- 1}$

Divide $d$ by $1$ .

$d = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$d = 1$