Question

select all the expressions that are equivalent to $a^x \cdot a^{-(x + y)}$: $\frac{1}{a^{x + y}}$, $(a^x)^{-(x + y)}$, $a^{x - (x + y)}$, $\frac{a^x}{a^y}$

Explanation:

Step1: Simplify the original expression

The original expression is \(a^{x} \cdot a^{-3x}\). Using the exponent rule \(a^{m} \cdot a^{n}=a^{m + n}\), we get \(a^{x+( - 3x)}=a^{-2x}\).

Step2: Analyze each option

• Option 1: \(\frac{1}{a^{2x}}\). Using the negative exponent rule \(a^{-n}=\frac{1}{a^{n}}\), since \(a^{-2x}=\frac{1}{a^{2x}}\), this option is equivalent.
• Option 2: \((a^{x})^{-3x}\). Using the exponent rule \((a^{m})^{n}=a^{m\times n}\), we get \(a^{x\times(-3x)} = a^{-3x^{2}}\), which is not equivalent to \(a^{-2x}\). Wait, maybe there is a typo? If it is \((a^{x})^{-2}\), then it would be \(a^{-2x}\), but as per the given option \((a^{x})^{-3x}\), it's not. Wait, maybe the original problem's second factor is \(a^{-2x}\)? Wait, the user's image shows the expression as \(a^{x}\cdot a^{-3x}\)? Wait, maybe I misread. Wait, let's re - check. Wait, the first option is \(\frac{1}{a^{2x}}\), the second is \((a^{x})^{-3x}\)? No, maybe it's \((a^{x})^{-2}\)? Wait, no, the user's image: let's assume that the correct simplification is for \(a^{x}\cdot a^{-2x}\) (maybe a typo in the exponent). Wait, no, let's do it correctly.

Wait, if the original expression is \(a^{x}\cdot a^{-2x}\), then \(a^{x-2x}=a^{-x}\)? No, no. Wait, let's take the first option \(\frac{1}{a^{2x}}=a^{-2x}\). The third option: \(a^{x - 3x}=a^{-2x}\) (if the original expression is \(a^{x}\cdot a^{-3x}\), then \(x+( - 3x)=-2x\), so \(a^{-2x}\), and the fourth option \(\frac{a^{x}}{a^{3x}}\). Using the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\), we get \(a^{x-3x}=a^{-2x}\).

Ah, I see, maybe the second option was a typo, but let's re - evaluate:

• For \(\frac{1}{a^{2x}}\): \(a^{-2x}=\frac{1}{a^{2x}}\), correct.
• For \((a^{x})^{-3x}\): No, that's \(a^{-3x^{2}}\), wrong. But if it's \((a^{x})^{-2}\), it would be \(a^{-2x}\), but as per the given, maybe it's a mistake. But the fourth option \(\frac{a^{x}}{a^{3x}}=a^{x - 3x}=a^{-2x}\), correct. The third option: if the expression is \(a^{x}\cdot a^{-3x}\), then \(a^{x-3x}=a^{-2x}\), so \(a^{-2x}\) is equivalent to \(\frac{1}{a^{2x}}\), \(a^{x-3x}\) (i.e., \(a^{-2x}\)), and \(\frac{a^{x}}{a^{3x}}\) (since \(\frac{a^{x}}{a^{3x}}=a^{x - 3x}=a^{-2x}\)). The second option \((a^{x})^{-3x}\) is incorrect. But maybe the second option is \((a^{x})^{-2}\), but as per the given, it's \((a^{x})^{-3x}\), which is wrong. But the first, third (if the exponent is \(x-3x\)) and fourth options are correct. Wait, the first option \(\frac{1}{a^{2x}}=a^{-2x}\), the third option \(a^{x-3x}=a^{-2x}\), the fourth option \(\frac{a^{x}}{a^{3x}}=a^{x - 3x}=a^{-2x}\). The second option \((a^{x})^{-3x}=a^{-3x^{2}}\) is wrong. So the correct options are \(\frac{1}{a^{2x}}\), \(a^{x - 3x}\) (i.e., \(a^{-2x}\)) and \(\frac{a^{x}}{a^{3x}}\).

Answer:

The equivalent expressions are \(\boldsymbol{\frac{1}{a^{2x}}}\), \(\boldsymbol{a^{x - 3x}}\) (or \(a^{-2x}\)) and \(\boldsymbol{\frac{a^{x}}{a^{3x}}}\). (Note: There might be a typo in the second option \((a^{x})^{-3x}\) as it is not equivalent, but the other three (assuming the third option is \(a^{x-3x}\)) are equivalent.)