Answer:
YES
Step-by-step explanation:
[tex]x^3\cdot x^{-7}=x^{3+(-7)}=x^{-4}\\\\\text{used}\ a^n\cdot a^m=a^{n+m}\\\\\text{Other way}\\\\a^{-n}=\dfrac{1}{a^n}\\\\x^3\cdot x^{-7}=x^3\cdot\dfrac{1}{x^7}=\dfrac{x^3}{x^7}=\dfrac{x\!\!\!\!\diagup^1\cdot x\!\!\!\!\diagup^1\cdot x\!\!\!\!\diagup^1}{x\!\!\!\!\diagup_1\cdot x\!\!\!\!\diagup_1 \cdot x\!\!\!\!\diagup_1 \cdot x\cdot x\cdot x\cdot x}=\dfrac{1}{x^4}=x^{-4}[/tex]
