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

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Answer 1 · 2015-08-17T12:52:41

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

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

So the expression can be written as:
$2^{x} = 2^{2 \cdot (x + 1)}$ $2^x=2^(2*(x+1))$

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

Now 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)?