How do you solve $2^{3 x - 2} = 4^{x}$ $2^(3x-2)=4^x$ ?

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Answer 1 · 2016-03-29T19:19:48

$x = 2$ $x=2$

$2^{3 x - 2} = 4^{x}$ $2^(3x-2) =4^x$

$2^{3 x - 2} = 2^{2 x}$ $color(red)(2)^(3x-2)=color(red)(2)^(2x)$

$equation has the same base value$ $"equation has the same base value"$

$3 x - 2 = 2 x$ $3x-2=2x$

$3 x - 2 x = 2$ $3x-2x=2$

$x = 2$ $x=2$

How do you solve 23x−2=4x2^(3x-2)=4^x?