If 27^{2x - 2} = \left(\frac{1}{3}\right)^{x - 8}, what is the value of x? — SAT

If $27^{2x - 2} = \left(\frac{1}{3}\right)^{x - 8}$ , what is the value of $x$ ?

Answer: 2

The value of

x

2

To solve the equation

27^{2x - 2} = \left(\frac{1}{3}\right)^{x - 8}

, express both sides with the base

3

. Since

27 = 3^3

and

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

, the equation becomes

(3^3)^{2x - 2} = (3^{-1})^{x - 8}

. Applying the exponent rule

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

yields

3^{6x - 6} = 3^{8 - x}

. Since the bases are now the same, set the exponents equal to each other:

6x - 6 = 8 - x

. Adding

x

to both sides gives

7x - 6 = 8

, and adding

6

to both sides gives

7x = 14

. Dividing by

7

results in

x = 2

Rewrite

27

and

\frac{1}{3}

as powers of

3

27 = 3^3

and

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

This allows both sides of the equation to be expressed with a common base of

3

so their exponents can be equated.

Substitute these bases back into the equation and apply exponent rules.

(3^3)^{2x - 2} = (3^{-1})^{x - 8} \implies 3^{3(2x - 2)} = 3^{-(x - 8)} \implies 3^{6x - 6} = 3^{8 - x}

Using the power rule

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

, multiply the outer exponent by the power of the base.

Equate the exponents and solve for

x

6x - 6 = 8 - x \implies 7x = 14 \implies x = 2

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

Solving exponential equations by finding a common base and applying exponent rules.

Question