How do you condense #1/3ln x + ln y#?

1 Answer
Apr 14, 2016

Answer:

#1/3lnx+lny=ln(yroot(3)x)#

Explanation:

We can condense using identities

#log_ap_1+log_ap_2+log_ap_3+..+log_ap_n=log_a(p_1*p_2*p_3*...*p_n)#

and #nlog_ap=log_ap^n# and

#1/mlog_ap=log_ap^(1/m)=log_aroot(m)p#

Using these #1/3lnx+lny=lnroot(3)x+lny=ln(yroot(3)x)#