If \frac{3^{2m + 1} \cdot 9^{m - 2}}{27^{m - 1}} = 81, what is the value of m? — State PSC Exam

If $\frac{3^{2m + 1} \cdot 9^{m - 2}}{27^{m - 1}} = 81$ , what is the value of $m$ ?

Answer: 4

The value of

m

is 4.

By converting all terms in the equation to base 3 and applying index rules (

a^x \cdot a^y = a^{x+y}

and

\frac{a^x}{a^y} = a^{x-y}

), the expression simplifies directly to

3^m = 3^4

, leading to

m = 4

Express all terms with a common base of 3

9^{m-2} = (3^2)^{m-2} = 3^{2m-4}

and

27^{m-1} = (3^3)^{m-1} = 3^{3m-3}

Rewriting powers with a single prime base allows for simplification using the laws of indices.

Simplify the numerator using the product rule of exponents

3^{2m+1} \cdot 3^{2m-4} = 3^{(2m+1) + (2m-4)} = 3^{4m-3}

According to the product rule

a^x \cdot a^y = a^{x+y}

, exponents with the same base are added during multiplication.

Apply the quotient rule of exponents to simplify the left-hand side

\frac{3^{4m-3}}{3^{3m-3}} = 3^{(4m-3) - (3m-3)} = 3^m

According to the quotient rule

\frac{a^x}{a^y} = a^{x-y}

, exponents are subtracted during division.

Express the right-hand side in base 3 and solve for

m

3^m = 81 = 3^4 \implies m = 4

Since the bases on both sides of the equation are equal, the exponents must also be equal.

Exponential Equations and Laws of Indices

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