Verified Answer :

  • Express both sides as powers of 6: \(\frac{1}{36} = 6^{-2}\) and \(216 = 6^3\).
  • Substitute these into the equation to get \((6^{-2})^n = 6^3\).
  • Simplify the left side to obtain \(6^{-2n} = 6^3\).
  • Solve the resulting linear equation \(-2n = 3\) to find \(n = \boxed{-\frac{3}{2}}\).

Explanation

  1. Analysis of the equation and base selection

The problem requires finding the value of \(n\) that satisfies the equation \(\left(\frac{1}{36}\right)^n = 216\). We observe that both \(\frac{1}{36}\) and 216 are powers of the same base, which is 6. Specifically, \(\frac{1}{36} = \frac{1}{6^2} = 6^{-2}\) and \(216 = 6 \times 6 \times 6 = 6^3\). Expressing both sides with the same base is the most efficient way to solve for the exponent.

  1. Rewriting the equation with a common base

We substitute the powers of 6 into the original equation:
\[(6^{-2})^n = 6^3\]
Using the exponent rule \((a^x)^y = a^{xy}\), we simplify the left side:
\[6^{-2n} = 6^3\]

  1. Solving for n

Since the bases are identical and the expression is equal, the exponents must also be equal. This gives us the simple linear equation:
\[-2n = 3\]
To solve for \(n\), divide both sides by -2:
\[n = -\frac{3}{2}\]
Thus, the value of \(n\) is \(-\frac{3}{2}\). Looking at the options provided, this corresponds to option (b).

Examples

In fields like chemistry, decay and growth are often modeled using exponential functions with specific bases. For instance, if a substance reduces to a fraction of its original amount over time, one might use an equation similar to this to find the time \(n\) (or growth rate) required for the substance to reach a certain level. Understanding these exponential relationships helps in calculating half-lives, compound interest, or population dynamics efficiently.

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