How do you condense #1/3(log_8y+2log_8(y+4))-log_8(y-1)#?

Question

1772 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-25T21:53:33

#log_8(root(3)(y(y+4)^2)/(y-1))#

Since #kloga=loga^k#, the expression is equivalent to:

#1/3(log_8y+log_8(y+4)^2)-log_8(y-1)#

Since #loga+logb=log(ab)#, you get:

#1/3log_8y(y+4)^2-log_8(y-1)#

#=log_8(y(y+4)^2)^(1/3)-log_8(y-1)#

Since #loga-logb=log(a/b)#, you get

#log_8((y(y+4)^2)^(1/3)/(y-1))#

that can be written as:

#log_8(root(3)(y(y+4)^2)/(y-1))#