Solve for y 125=(1/25)^(y-1)

$125 = {(\frac{1}{25})}^{y - 1}$

Rewrite the equation as ${(\frac{1}{25})}^{y - 1} = 125$ .

${(\frac{1}{25})}^{y - 1} = 125$

Apply the product rule to $\frac{1}{25}$ .

$\frac{1^{y - 1}}{25^{y - 1}} = 125$

One to any power is one.

$\frac{1}{25^{y - 1}} = 125$

Move $25^{y - 1}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$25^{- (y - 1)} = 125$

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

$5^{2 (- (y - 1))} = 5^{3}$

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

$2 (- (y - 1)) = 3$

Solve for $y$ .

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Simplify.

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Apply the distributive property.

$2 (- y - - 1) = 3$

Multiply $- 1$ by $- 1$ .

$2 (- y + 1) = 3$

Divide each term by $- 2$ and simplify.

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Divide each term in $2 (- y + 1) = 3$ by $- 2$ .

$\frac{2 (- y + 1)}{- 2} = \frac{3}{- 2}$

Simplify $\frac{2 (- y + 1)}{- 2}$ .

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Simplify the expression.

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Move the negative one from the denominator of $\frac{- y + 1}{- 1}$ .

$- 1 \cdot (- y + 1) = \frac{3}{- 2}$

Rewrite $- 1 \cdot (- y + 1)$ as $- (- y + 1)$ .

$- (- y + 1) = \frac{3}{- 2}$

Apply the distributive property.

$- - y - 1 \cdot 1 = \frac{3}{- 2}$

Multiply $- - y$ .

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Multiply $- 1$ by $- 1$ .

$1 y - 1 \cdot 1 = \frac{3}{- 2}$

Multiply $y$ by $1$ .

$y - 1 \cdot 1 = \frac{3}{- 2}$

Multiply $- 1$ by $1$ .

$y - 1 = \frac{3}{- 2}$

Move the negative in front of the fraction.

$y - 1 = - \frac{3}{2}$

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

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Add $1$ to both sides of the equation.

$y = - \frac{3}{2} + 1$

Write $1$ as a fraction with a common denominator.

$y = - \frac{3}{2} + \frac{2}{2}$

Combine the numerators over the common denominator.

$y = \frac{- 3 + 2}{2}$

Add $- 3$ and $2$ .

$y = \frac{- 1}{2}$

Move the negative in front of the fraction.

$y = - \frac{1}{2}$

The result can be shown in multiple forms.

Exact Form:

$y = - \frac{1}{2}$

Decimal Form:

$y = - 0.5$

Solve for y 125=(1/25)^(y-1)

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