How do you solve #4^(x + 7) = 6^(x – 1)#?

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Answer 1 · 2016-04-10T13:27:19

#" "x=(8ln(4))/(ln(6)-ln(4))+1#

#" "x~~28.352# to 3 decimal places

Take logs

#" "ln(4^(x+7))=ln(6^(x-1))#

#" "=>(x+7)ln(4)=(x-1)ln(6)#

#" "=>(x+7)/(x-1)=ln(6)/ln(4)#.............................(1)

But #x-7# can be written as #x-1+8#

Write equation (1) as

#" "=>(x-1+8)/(x-1)=ln(6)/ln(4)#

#" "=>(x-1)/(x-1)+8/(x-1)=ln(6)/ln(4)#

#" "1+8/(x-1)=ln(6)/ln(4)#

#" "8/(x-1)=ln(6)/ln(4)-1#

#" "8/(x-1)=(ln(6)-ln(4))/ln(4)#

#" "x-1= (8ln(4))/(ln(6)-ln(4))#

#" "x=(8ln(4))/(ln(6)-ln(4))+1#

#" "x~~28.352# to 3 decimal places