Answer:
[tex]5^{-1}a^2c^{-3}[/tex]
Step-by-step explanation:
[tex]We\ are\ given,\\\frac{(5a^3c)}{(25ac^4)}\\=\frac{a^3c}{5ac^4} [Cancelling\ 5\ from\ the\ numerator\ and\ denominator]\\Now,\\We\ may\ use\ one\ of\ the\ laws\ of\ exponents\ below:[/tex]
[tex]* For\ any\ real\ number\ n, \frac{n^a}{n^b} =n^{(a-b)}\\ * For\ any\ real\ number\ n, \frac{1}{n}=n^{-1}[/tex]
[tex]Hence,\\\frac{1}{5}a^{(3-1)}*c^{(1-4)}\\= \frac{1}{5}a^{(2)}*c^{(-3)}[ Evaluating]\\= \frac{1}{5}a^{(2)}c^{(-3)}\\=5^{-1}a^2c^{-3}[/tex]
