Question

If \( x = \left(\dfrac{3}{2}\right)^{2}\left(\dfrac{2}{3}\right)^{-4} \), the value of \( x^{-2} \) is

• A. \( \left(\dfrac{2}{3}\right)^{12} \)
• B. \( \left(\dfrac{3}{2}\right)^{12} \)
• C. \( \left(\dfrac{6}{5}\right)^{-12} \)
• D. \( \left(\dfrac{5}{6}\right)^{-12} \)

Hint:

The bases 3/2 and 2/3 are reciprocals. Flip the second one so both factors share the base 3/2, add the exponents to get x, then use the outer power of minus 2.

Answer 1

The Correct Option is A

Step 1: Write everything with prime bases 2 and 3.
Instead of juggling fractions, split each fraction into powers of 2 and 3.
\[ \left(\frac{3}{2}\right)^{2} = \frac{3^{2}}{2^{2}} = 3^{2} \cdot 2^{-2} \]
\[ \left(\frac{2}{3}\right)^{-4} = \frac{2^{-4}}{3^{-4}} = 2^{-4} \cdot 3^{4} \]

Step 2: Multiply and collect each prime.
For base 3: $3^{2} \cdot 3^{4} = 3^{6}$.
For base 2: $2^{-2} \cdot 2^{-4} = 2^{-6}$.
So \[ x = 3^{6} \cdot 2^{-6} \]

Step 3: Apply the outer exponent.
Multiply each exponent by $-2$:
\[ x^{-2} = 3^{6 \times (-2)} \cdot 2^{-6 \times (-2)} = 3^{-12} \cdot 2^{12} \]

Step 4: Put it back as a single fraction.
\[ 2^{12} \cdot 3^{-12} = \frac{2^{12}}{3^{12}} = \left(\frac{2}{3}\right)^{12} \]
As a number this is $\dfrac{4096}{531441}$, a value less than 1, which makes sense because $x$ is bigger than 1 and we took a negative power of it.

Step 5: Rule out the rest.
$\left(\frac{3}{2}\right)^{12}$ is about 129.7, far bigger than 1, so it cannot equal $x^{-2}$.
$\left(\frac{6}{5}\right)^{-12}$ and $\left(\frac{5}{6}\right)^{-12}$ both bring in the prime 5, and no 5 exists anywhere in $x$. Their prime factorisation can never match $2^{12} \cdot 3^{-12}$.

Final Answer:
Breaking into primes gives the same result, option (A). \[ \boxed{\left(\frac{2}{3}\right)^{12}} \]