How do you condense #2/3ln y + ln t – ln y -1/2 ln t#?

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Answer 1 · 2016-09-02T01:27:13

#2/3lny+lnt-lny-1/2lnt=ln(sqrtt/(root(3)y))#

#2/3lny+lnt-lny-1/2lnt#

= #lny(2/3-1)+lnt(1-1/2)#

= #-1/3lny+1/2lnt#

Now as #1/mlna=lna^(1/m)=lnroot(m)a# and #-lnb=ln(1/b)#

#-1/3lny+1/2lnt#

= #ln(1/root(3)y)+lnsqrtt#

= #ln(sqrtt/(root(3)y))#