Solve the equation:
$216 = \left(\frac{1}{36}\right)^{x-1}$
We are given the exponential equation \(216 = \left(\frac{1}{36}\right)^{x-1}\). The goal is to solve for \(x\). We observe that both \(216\) and \(\frac{1}{36}\) are powers of \(6\). Specifically, \(216 = 6^3\) and \(36 = 6^2\), which implies \(\frac{1}{36} = 6^{-2}\). This allows us to express both sides of the equation using a common base of \(6\).
Starting with the given equation \(216 = \left(\frac{1}{36}\right)^{x-1}\), we substitute the powers of \(6\):
\[6^3 = (6^{-2})^{x-1}\]
Using the power of a power rule \((a^m)^n = a^{mn}\), we simplify the right side of the equation:
\[6^3 = 6^{-2(x-1)}\]
\[6^3 = 6^{-2x+2}\]
Since the bases are equal, we can set the exponents equal to each other:
\[3 = -2x + 2\]
Now we solve the resulting linear equation for \(x\):
\[3 = -2x + 2\]
Subtracting 2 from both sides, we get:
\[1 = -2x\]
Dividing by \(-2\), we find:
\[x = -\frac{1}{2}\]
Thus, the correct value for \(x\) is \(-\frac{1}{2}\). This matches the first option provided in the problem description.
Exponential equations appear in many real-world scenarios, such as calculating compound interest or modeling population growth. For instance, if you have a financial investment that grows exponentially, you might need to determine the time \(t\) required to reach a specific value, which involves solving an equation similar to the one presented here by converting both sides to a common base or using logarithms. Understanding these techniques allows for precise predictions in economics and science.