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

Question

1313 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-23T10:49:12

#((16b^(-2)c)(25b^4c^(-5)))/((15b^5c^(-1))(8b^(-7)c^(-2)))=(10b^4)/(3c)#

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

we will use identities #a^mxxa^n=a^(m+n)#, #a^m/a^n=a^(m-n)#, #a^(-m)=1/a^m# and #1/a^(-m)=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)#

= #(2xx2xx2xx2xx5xx5)/(3xx5xx2xx2xx2)xx(c^((1+2+1))b^((4+7)))/(b^((2+5))c^5)#

= #(2xx5)/(3)xx(c^4b^11)/(b^7c^5)=10/3xxb^((11-7))/c^((5-4))#

= #(10b^4)/(3c)#