If 3^{2x - 1} = 81 \cdot 9^{y + 2}, which of the following equations correctly expresses x… — SAT

If $3^{2x - 1} = 81 \cdot 9^{y + 2}$ , which of the following equations correctly expresses $x$ in terms of $y$ ?

x = y + \frac{9}{2}

The correct equation is found by expressing

81

3^4

and

9^{y+2}

3^{2y+4}

. Applying the product rule for exponents, the right side becomes

3^{2y+8}

. Since the bases are the same, equating the exponents gives

2x - 1 = 2y + 8

. Solving for

x

yields

x = y + \frac{9}{2}

Rewrite all parts of the equation using a common base of

3

Since

81 = 3^4

and

9 = 3^2

, the term

9^{y+2}

becomes

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

. The equation can be rewritten as

3^{2x - 1} = 3^4 \cdot 3^{2y + 4}

Expressing all exponential terms with the same base allows the use of exponent rules to simplify the equation.

Simplify the product on the right side of the equation using the product rule for exponents,

a^m \cdot a^n = a^{m+n}

3^{2x - 1} = 3^{4 + (2y + 4)}

, which simplifies to

3^{2x - 1} = 3^{2y + 8}

Adding the exponents of terms with a common base simplifies the right side into a single exponential expression.

Set the exponents equal to each other.

2x - 1 = 2y + 8

If two exponential expressions with the same positive base (other than

1

) are equal, their exponents must be equal.

Solve for

x

in terms of

y

Add

1

to both sides to get

2x = 2y + 9

, then divide by

2

to obtain

x = y + \frac{9}{2}

This isolates the variable

x

to express it as a function of

y

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

Question