If $\left(\frac{9}{4}\right)^x \cdot \left(\frac{8}{27}\right)^{x-1} = \frac{2}{3}$, then the value of $x$ is

### Concept & Formula When an equation features fractions as bases, the objective is to invert or manipulate them until they share identical fractional bases. This relies heavily on recognizing perfect squares and cubes of small fractions, and using negative exponents to flip fractions. The pivotal rules are: $$\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$$ $$\left(x^m\right)^n = x^{mn}$$ ### Step-by-Step Solution * **Given:** $$\left(\frac{9}{4}\right)^x \cdot \left(\frac{8}{27}\right)^{x-1} = \frac{2}{3}$$ * **Calculation:** Recognize that the fractions are related to base $\frac{3}{2}$ and $\frac{2}{3}$. Let us standardize to base $\frac{3}{2}$. Convert the first term: $\frac{9}{4} = \left(\frac{3}{2}\right)^2$ So, $\left(\frac{9}{4}\right)^x = \left(\left(\frac{3}{2}\right)^2\right)^x = \left(\frac{3}{2}\right)^{2x}$ Convert the second term: $\frac{8}{27} = \left(\frac{2}{3}\right)^3$ To match our base of $\frac{3}{2}$, flip it and negate the exponent: $\left(\frac{2}{3}\right)^3 = \left(\frac{3}{2}\right)^{-3}$ So, $\left(\frac{8}{27}\right)^{x-1} = \left(\left(\frac{3}{2}\right)^{-3}\right)^{x-1} = \left(\frac{3}{2}\right)^{-3(x-1)} = \left(\frac{3}{2}\right)^{-3x+3}$ Convert the right-hand side: $\frac{2}{3} = \left(\frac{3}{2}\right)^{-1}$ Substitute everything back into the equation: $$\left(\frac{3}{2}\right)^{2x} \cdot \left(\frac{3}{2}\right)^{-3x+3} = \left(\frac{3}{2}\right)^{-1}$$ Combine the LHS exponents using addition: $$\left(\frac{3}{2}\right)^{2x - 3x + 3} = \left(\frac{3}{2}\right)^{-1}$$ $$\left(\frac{3}{2}\right)^{-x + 3} = \left(\frac{3}{2}\right)^{-1}$$ Equate the exponents: $$-x + 3 = -1$$ $$-x = -4$$ $$x = 4$$ ### Exam Strategy & Shortcut **Option Verification:** For fractional equations that look algebraically messy, plugging in the options can be faster. Try Option (d) where $x = 4$: LHS $= \left(\frac{9}{4}\right)^4 \cdot \left(\frac{8}{27}\right)^3$ $= \frac{3^8}{2^8} \cdot \frac{2^9}{3^9}$ $= \frac{3^8 \cdot 2^9}{2^8 \cdot 3^9} = \frac{2^1}{3^1} = \frac{2}{3}$ This perfectly matches the RHS, confirming the answer instantly without full algebra. ### Common Pitfall Failing to distribute the negative sign properly when multiplying the exponents is the most frequent error. Writing $-3(x-1)$ as $-3x-3$ instead of the correct $-3x+3$ leads to a completely incorrect final value for $x$. ### Final Answer **Therefore, the correct answer is 4.**