Solve:
$243^{-y}=\left(\frac{1}{243}\right)^{3 y} \cdot 9^{-2 y}$
The given equation is \(243^{-y} = \left(\frac{1}{243}\right)^{3y} \cdot 9^{-2y}\). To solve for \(y\), we express all bases as powers of 3. Since \(243 = 3^5\) and \(9 = 3^2\), we can substitute these into the equation. We also know that \(\frac{1}{243} = 243^{-1} = (3^5)^{-1} = 3^{-5}\).
Substituting the base values into the equation, we get:
\[ (3^5)^{-y} = (3^{-5})^{3y} \cdot (3^2)^{-2y} \]
Applying the power of a power rule \((a^m)^n = a^{mn}\):
\[ 3^{-5y} = 3^{-15y} \cdot 3^{-4y} \]
Using the product rule for exponents \(a^m \cdot a^n = a^{m+n}\):
\[ 3^{-5y} = 3^{-15y + (-4y)} = 3^{-19y} \]
Since the bases on both sides are equal and not equal to 1 or -1, we can equate the exponents:
\[ -5y = -19y \]
To solve this linear equation, we add \(19y\) to both sides:
\[ -5y + 19y = 0 \]
\[ 14y = 0 \]
\[ y = 0 \]
Thus, the value of \(y\) that satisfies the equation is \(0\). We can verify this: \(243^0 = 1\) and \((1/243)^0 \cdot 9^0 = 1 \cdot 1 = 1\), so the solution is correct.
In physics, such equations often appear when dealing with radioactive decay or exponential growth where quantities are defined by specific bases. For example, if you are modeling the cooling of an object or the degradation of a compound, setting rates equal to one another helps determine the exact time \(t\) or parameter \(y\) at which two processes reach equilibrium. Mastering exponent laws is essential for solving these types of problems in chemistry, biology, and engineering.