How do you rewrite #2lnx+1/2lnu# as a single log?

1 Answer
Jul 23, 2015

Answer:

I found: #ln(x^2sqrt(u))#

Explanation:

Here you can use two properties of the logs:
1] #alogx=logx^a#
2] #log_bx+log_by=log_b(x*y)# (same base!)

So, in your case you get:
#lnx^2+lnu^(1/2)=lnx^2+lnsqrt(u)=# same base:
#=ln(x^2sqrt(u))#