Answer:
The expression without a negative exponent will be:
[tex]\frac{8x^{-4}y^{-8}}{-2xy^5}=-\frac{4}{x^5y^{13}}[/tex]
Step-by-step explanation:
Given the function
[tex]\frac{8x^{-4}\:y^{-8}}{-2xy^5}[/tex]
[tex]\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{-b}=-\frac{a}{b}[/tex]
[tex]\frac{8x^{-4}\:y^{-8}}{-2xy^5}=-\frac{8x^{-4}y^{-8}}{2xy^5}\\[/tex]
[tex]\mathrm{Divide\:the\:numbers:}\:\frac{8}{2}=4[/tex]
[tex]=-\frac{4x^{-4}y^{-8}}{xy^5}[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=\frac{1}{x^{b-a}}[/tex]
[tex]=-\frac{4y^{-8}}{x^5y^5}[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=\frac{1}{x^{b-a}}[/tex]
i.e. [tex]\frac{y^{-8}}{y^5}=\frac{1}{y^{5-\left(-8\right)}}[/tex]
so the expression becomes
[tex]\frac{8x^{-4}\:y^{-8}}{-2xy^5}=-\frac{4}{x^5y^{5-\left(-8\right)}}[/tex]
[tex]=-\frac{4}{x^5y^{13}}[/tex]
Therefore, the expression without a negative exponent will be:
[tex]\frac{8x^{-4}y^{-8}}{-2xy^5}=-\frac{4}{x^5y^{13}}[/tex]
