Simplify #5lnx +21lnx^3-3lnx^3+lnx^(1/2)#?

Question

1661 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-03T13:06:43

#5lnx +21lnx^3-3lnx^3+lnx^(1/2)=119/2lnx#

We can use the identities #lna+lnb-lnc=ln((ab)/c)# and #lna^b=blna#.

Hence,

#5lnx +21lnx^3-3lnx^3+lnx^(1/2)#

= #5lnx +21xx3lnx-3xx3lnx+1/2lnx#

= #5lnx +63lnx-9lnx+1/2lnx#

= #lnx(5+63-9+1/2)#

= #lnx(59+1/2)#

= #119/2lnx#