If 9^{2x + 1} = \left(\frac{1}{27}\right)^{x - 2}, what is the value of x? — SAT

If $9^{2x + 1} = \left(\frac{1}{27}\right)^{x - 2}$ , what is the value of $x$ ?

\frac{4}{7}

The correct answer is

\frac{4}{7}

. By rewriting both sides of the equation with a common base of

3

, the equation becomes

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

. Applying the power of a power rule gives

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

. Since the bases are equal, the exponents must be equal, giving

4x + 2 = -3x + 6

. Solving for

x

results in

7x = 4

, which gives

x = \frac{4}{7}

Rewrite each side of the equation with a common base of

3

9 = 3^2

and

\frac{1}{27} = 3^{-3}

, so the equation becomes

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

Before solving an exponential equation, it is helpful to express the bases in terms of their common prime base.

Apply the exponent power rule

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

to simplify the exponents.

3^{2(2x + 1)} = 3^{-3(x - 2)} \implies 3^{4x + 2} = 3^{-3x + 6}

Simplifying the expressions on both sides allows for equating the exponents directly.

Set the exponents equal to each other and solve the resulting linear equation.

4x + 2 = -3x + 6 \implies 7x = 4 \implies x = \frac{4}{7}

Since the bases are equal, the powers can only be equal if their exponents are equal.

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

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