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

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