How do you solve #14^ { 4x + 1} = 5^ { x - 2}#?

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Answer 1 · 2018-01-08T20:01:04

#x=-(ln14+2ln5)/(4ln14-ln5)#

#14^(4x+1)=5^(x-2)#

#ln14^(4x+1)=ln5^(x-2)#

#(4x+1)ln14=(x-2)ln5#

#4xln14+ln14=xln5-2ln5#

#4xln14-xln5=-ln14-2ln5#

#x(4ln14-ln5)=-ln14-2ln5#

#x=-(ln14+2ln5)/(4ln14-ln5)#

Answer 2 · 2018-01-08T20:01:49

#x=(log(5)xx2-log(14))/(log(14)xx4-log(5))~~0,06480742301227#

#14^(4x+1)=5^(x−2)quadquadquadquad//xxlog#

#log14^(4x+1)=log5^(x−2)#

#log(14)xx(4x+1)=log(5)xx(x−2)#

#log(14)xx(4x+1)-log(5)xx(x−2)=0#

#log(14)xx4x+log(14)-log(5)xxx−log(5)xx2=0#

#log(14)xx4x-log(5)xxx=log(5)xx2-log(14)#

Taking out x

#x xxcolor(grey)((log(14)xx4-log(5)))=log(5)xx2-log(14)#

#x=(log(5)xx2-log(14))/(log(14)xx4-log(5))~~0,06480742301227#