# How do you simplify ((16b^-2c)(25b^4c^-5))/((15b^5c^-1)(8b^-7c^-2))?

Jun 23, 2016

$\frac{\left(16 {b}^{- 2} c\right) \left(25 {b}^{4} {c}^{- 5}\right)}{\left(15 {b}^{5} {c}^{- 1}\right) \left(8 {b}^{- 7} {c}^{- 2}\right)} = \frac{10 {b}^{4}}{3 c}$

#### Explanation:

For simplifying ((16b^(-2)c)(25b^4c^(-5)))/((15b^5c^(-1))(8b^(-7)c^(-2))

we will use identities ${a}^{m} \times {a}^{n} = {a}^{m + n}$, ${a}^{m} / {a}^{n} = {a}^{m - n}$, ${a}^{- m} = \frac{1}{a} ^ m$ and $\frac{1}{a} ^ \left(- m\right) = {a}^{m}$

Hence ((16b^(-2)c)(25b^4c^(-5)))/((15b^5c^(-1))(8b^(-7)c^(-2))

= ((16c)(25b^4)b^7c^2*c)/(b^2c^5(15b^5)(8)

= (16xx25)/(15xx8)xx(c^1b^4b^7c^2*c^1)/(b^2c^5(b^5)

= $\frac{2 \times 2 \times 2 \times 2 \times 5 \times 5}{3 \times 5 \times 2 \times 2 \times 2} \times \frac{{c}^{\left(1 + 2 + 1\right)} {b}^{\left(4 + 7\right)}}{{b}^{\left(2 + 5\right)} {c}^{5}}$

= $\frac{2 \times 5}{3} \times \frac{{c}^{4} {b}^{11}}{{b}^{7} {c}^{5}} = \frac{10}{3} \times {b}^{\left(11 - 7\right)} / {c}^{\left(5 - 4\right)}$

= $\frac{10 {b}^{4}}{3 c}$