If 81^{y-2} = 27^{y+1}, what is the value of y? — SAT

If $81^{y-2} = 27^{y+1}$ , what is the value of $y$ ?

Answer: 11

The correct answer is 11. By expressing both sides of the equation with the common base of 3, the equation simplifies from

81^{y-2} = 27^{y+1}

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

. Applying the power of a power rule gives

3^{4y-8} = 3^{3y+3}

. Equating the exponents yields

4y - 8 = 3y + 3

, which solves to

y = 11

Express both bases as powers of 3

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

To solve an exponential equation algebraically, it is helpful to express both sides using a common base.

Apply the power of a power exponent rule

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

The rule

(a^m)^n = a^{mn}

allows us to simplify the exponent expressions by multiplying the exponents.

Equate the exponents

4(y-2) = 3(y+1)

Since the bases are equal, the exponents must be equal for the equation to hold true.

Distribute the coefficients

4y - 8 = 3y + 3

Expanding the linear expressions prepares the equation for isolation of the variable.

Solve the linear equation for

y

y = 11

Subtract

3y

from both sides to get

y - 8 = 3

, then add 8 to both sides to isolate

y

Solving exponential equations by expressing bases in terms of a common base and equating exponents.

Alternatively, substitute the value of 11 back into the original equation to verify that both sides are equal:

81^{11-2} = 81^9 = (3^4)^9 = 3^{36}

and

27^{11+1} = 27^{12} = (3^3)^{12} = 3^{36}

Estimated Time:1m 30s

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