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

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Answer 1 · 2015-09-09T06:25:43

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

$2^{x} = 4^{x + 1}$ $2^x = 4^(x+1)$

We know that $4 = 2^{2}$ $4 = 2^2$

So,
$2^{x} = 2^{2 (x + 1)}$ $2^x = 2^(2(x+1))$

$2^{x} = 2^{(2 x + 2)}$ $2^color(blue)(x) = 2^color(blue)((2x +2 )$

At this stage as the bases are equal we equate the exponents to find $x$ $x$

$x = 2 x + 2$ $x = 2x+2$
$- 2 = 2 x - x$ $-2 = 2x-x$
$x = - 2$ $color(blue)(x = -2$

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