Certainly! Let's break down the equation [tex]\(243^{-y} = \left(\frac{1}{243}\right)^{3y} \cdot 9^{-2y}\)[/tex] step-by-step.
1. Rewrite the equation with common bases:
- Notice that [tex]\(243 = 3^5\)[/tex] and [tex]\(9 = 3^2\)[/tex].
- Therefore, we can rewrite each term in the equation using base 3.
2. Convert to base 3:
- [tex]\(243^{-y} = (3^5)^{-y} = 3^{-5y}\)[/tex]
- [tex]\(\left(\frac{1}{243}\right)^{3y} = (243^{-1})^{3y} = (3^{-5})^{3y} = 3^{-15y}\)[/tex]
- [tex]\(9^{-2y} = (3^2)^{-2y} = 3^{-4y}\)[/tex]
3. Substitute these back into the original equation:
- [tex]\(3^{-5y} = 3^{-15y} \cdot 3^{-4y}\)[/tex]
4. Simplify the right-hand side:
- When multiplying powers with the same base, add the exponents: [tex]\(3^{-15y} \cdot 3^{-4y} = 3^{-15y + (-4y)} = 3^{-19y}\)[/tex]
5. Equate the exponents:
- We now have [tex]\(3^{-5y} = 3^{-19y}\)[/tex]
- Since the bases are the same, we can set the exponents equal to each other: [tex]\(-5y = -19y\)[/tex]
6. Solve for [tex]\(y\)[/tex]:
- Combine like terms: [tex]\(-5y + 19y = 0\)[/tex]
- Simplify: [tex]\(14y = 0\)[/tex]
- Divide by 14: [tex]\(y = 0\)[/tex]
Therefore, the solution is [tex]\(y = 0\)[/tex].
