If 9^{y - 1} \cdot 3^{2y} = 27^{y + 2}, what is the value of y? — SAT

If $9^{y - 1} \cdot 3^{2y} = 27^{y + 2}$ , what is the value of $y$ ?

To solve the equation, we rewrite each term using the common base of

3

. Since

9 = 3^2

and

27 = 3^3

, the equation

9^{y - 1} \cdot 3^{2y} = 27^{y + 2}

can be written as

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

. Applying the power of a power rule, we get

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

, which simplifies to

3^{2y-2} \cdot 3^{2y} = 3^{3y+6}

. Using the product rule of exponents to combine the left side gives

3^{2y-2+2y} = 3^{4y-2} = 3^{3y+6}

. Setting the exponents equal to each other gives the linear equation

4y - 2 = 3y + 6

. Solving for

y

yields

y = 8

Express all bases in the equation as powers of 3.

9^{y-1} = (3^2)^{y-1} = 3^{2y-2}

and

27^{y+2} = (3^3)^{y+2} = 3^{3y+6}

. The original equation becomes

3^{2y-2} \cdot 3^{2y} = 3^{3y+6}

Having a common base allows us to combine exponents using the properties of exponents.

Simplify the left side of the equation by adding the exponents.

3^{(2y-2) + 2y} = 3^{4y-2}

, so the equation is

3^{4y-2} = 3^{3y+6}

When multiplying exponential expressions with the same base, we add their exponents.

Set the exponents equal to each other and solve for

y

4y - 2 = 3y + 6 \implies y - 2 = 6 \implies y = 8

Since the bases are equal and positive (and not equal to 1), their exponents must be equal.

Solving exponential equations by expressing terms with a common base and applying exponent laws.

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