If 3^{x+1} = k, where k > 0, which of the following is equivalent to 27^{x-1}? — SAT

If $3^{x+1} = k$ , where $k > 0$ , which of the following is equivalent to $27^{x-1}$ ?

The expression \frac{k^3}{729}

To express

27^{x-1}

in terms of

k

, we can rewrite both the given equation and the target expression using base

3

. The given equation

3^{x+1} = k

can be adjusted to find the value of

3^{x-1}

by dividing both sides by

3^2

, which yields

3^{x-1} = \frac{k}{9}

. The target expression

27^{x-1}

can be rewritten as

(3^3)^{x-1} = (3^{x-1})^3

. Substituting

3^{x-1} = \frac{k}{9}

into this expression gives

\left(\frac{k}{9}\right)^3 = \frac{k^3}{729}

Express the given equation in terms of a simpler base

3

exponent.

3^{x-1} = \frac{k}{9}

We start with the given equation

3^{x+1} = k

. To relate this to the exponent

x-1

, we divide both sides of the equation by

3^2 = 9

3^{x-1} = \frac{3^{x+1}}{3^2} = \frac{k}{9}

Rewrite the target expression

27^{x-1}

with base

3

27^{x-1} = (3^{x-1})^3

Since

27 = 3^3

, we can rewrite the target expression as

27^{x-1} = (3^3)^{x-1} = 3^{3(x-1)} = (3^{x-1})^3

Substitute the expression for

3^{x-1}

from Step 1 into the expression from Step 2.

\frac{k^3}{729}

Substituting

3^{x-1} = \frac{k}{9}

into

(3^{x-1})^3

gives

\left(\frac{k}{9}\right)^3 = \frac{k^3}{9^3} = \frac{k^3}{729}

Manipulating exponential equations by expressing terms with a common base and applying exponent rules.

Question