If \left( \frac{16}{81} \right)^{x-1} \times \left( \frac{27}{8} \right)^{x+2} = \frac{9}{… — JAMB UTME

If $\left( \frac{16}{81} \right)^{x-1} \times \left( \frac{27}{8} \right)^{x+2} = \frac{9}{4}$ , find the value of $x$ .

The value of

x

8

By writing

\frac{16}{81}

\left(\frac{2}{3}\right)^4

\frac{27}{8}

\left(\frac{2}{3}\right)^{-3}

, and

\frac{9}{4}

\left(\frac{2}{3}\right)^{-2}

, the equation simplifies via exponent addition to

\left(\frac{2}{3}\right)^{4x - 4 - 3x - 6} = \left(\frac{2}{3}\right)^{-2}

. Equating indices gives

x - 10 = -2

, which yields the correct solution

x = 8

Express all fractional terms with a common base of

\frac{2}{3}

\frac{16}{81} = \left(\frac{2}{3}\right)^4

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

, and

\frac{9}{4} = \left(\frac{3}{2}\right)^2 = \left(\frac{2}{3}\right)^{-2}

Converting all terms to a single common base allows exponents to be combined using the laws of indices.

Substitute the common base expressions back into the original equation

\left( \left(\frac{2}{3}\right)^4 \right)^{x-1} \times \left( \left(\frac{2}{3}\right)^{-3} \right)^{x+2} = \left(\frac{2}{3}\right)^{-2}

Applying the power of a power law

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

to simplify each term.

Apply the power law and multiplication law of indices

\left(\frac{2}{3}\right)^{4(x-1)} \times \left(\frac{2}{3}\right)^{-3(x+2)} = \left(\frac{2}{3}\right)^{-2} \implies \left(\frac{2}{3}\right)^{4(x-1) - 3(x+2)} = \left(\frac{2}{3}\right)^{-2}

When multiplying exponential terms with identical bases, add their exponents:

a^m \times a^n = a^{m+n}

Equate the exponents and solve for

x

4(x-1) - 3(x+2) = -2 \implies 4x - 4 - 3x - 6 = -2 \implies x - 10 = -2 \implies x = 8

Since the bases on both sides are equal and non-zero, their indices must be equal.

Laws of Indices: Base Conversion and Exponential Equations

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