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

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Answer 1 · 2015-06-17T04:23:43

$x = - 2$ $color(red)( x = -2$

$2^{x} = 2^{2 . (x + 1)}$ $2^color(red)(x)=2^(color(red)(2.(x+1)$
Here as bases are equal we can equate the respective powers with each other:

$x = 2 (x + 1)$ $x = 2(x+1)$
$x = 2 x + 2$ $x = 2x +2$
$x = - 2$ $color(red)( x = -2$

How do you solve 2^x=4^(x+1)2x=4x+12^x=4^(x+1)?