If $\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1$, find the value of x.

Step 1: Analyze the question We need to find x such that the expression equals 1. The trick is to write every number as a power of 3, then compare the exponents.

Step 2: Concept Laws of indices:

• (a^m)^n = a^(mn)
• a^m × a^n = a^(m+n)
• a^m ÷ a^n = a^(m−n)
• If a^p = a^q, then p = q

Step 3: Solution Rewrite each base as a power of 3:

• 27^x = (3³)^x = 3^(3x)
• 9^(2x) = (3²)^(2x) = 3^(4x)

Substitute:

3^(3x) × 3^(1−x) / 3^(4x) = 1

Combine the exponents on top:

3^(3x + 1 − x) / 3^(4x) = 1

3^(2x + 1) / 3^(4x) = 1

3^(2x + 1 − 4x) = 3^0

So 1 − 2x = 0, which gives:

x = 1/2

Check: With x = 1/2:

• 27^(1/2) = 3^(3/2)
• 3^(1 − 1/2) = 3^(1/2)
• 9^(1) = 9 = 3²

Numerator: 3^(3/2 + 1/2) = 3² = 9. Then 9/9 = 1 ✓

Answer: x = 1/2, which is option B.

Tip: In index equations, first convert everything to the same base (here 3). Then the equation reduces to a simple one in the exponents. Also remember that anything equal to 1 can be written as base^0.