lf `(3a)^log3= (4b)^log4` and `4^loga=3^logb` ,then a+b=

1.	lf `(3a)^log3= (4b)^log4` and `4^loga=3^logb` ,then a+b=
Answer» `(3a)^(log3)=(4b)^(log4)` `log[log(3a)]=log4[log(4b)]` `log3[log3+loga]+log4[log4+logb]-(1)` `4^(loga)=3^(logb)` `logalog4=logblog3` `loga=(logblog3)/(log4)-(2)` `log3[log3+(logblog3)/(log4)]=log4[log4+logb]` `(log3)^2[1+(logb)/(loga)]=log4[log4+logb]` `(log)^3(log4+logb)=(log4)^2[log4+logb]` `log4+logb=0` `b=1/4` `loga=(log(1/4)log3)/(log4)` `=(-log4log3)/(log4)` `a=1/3` `a+b=1/3+1/4=(4+3)/12=7/12`.

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lf `(3a)^log3= (4b)^log4` and `4^loga=3^logb` ,then a+b=

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