How do you solve $2^(x+1) = 3(4^x)$ ?

Question

1153 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-12T09:22:12

$x = 1-(log 3)/log 2$ .

Use $(a^m)^n=a^(mn) and a^m/a^n=a^(m-n)$ .

$2^(x+1)=3(4^x)=3(2^2)^x=3(2^(2x))$
So, 2^(x+1-2x)=3 or 2^(1-x)+3.
Equating logarithms, $(1-x) log 2=log 3$ .
$x=1-(log 3)/log 2$ ..

How do you solve 2^(x+1) = 3(4^x)2^(x+1) = 3(4^x)?