How do you solve $9^{x} = 3^{x - 2}$ $9^x=3^(x-2)$ ?

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Answer 1 · 2018-04-03T13:25:31

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

We got: $9^{x} = 3^{x - 2}$ $9^x=3^(x-2)$

And so:

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

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

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

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

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

How do you solve 9^x=3^(x-2)9x=3x−29^x=3^(x-2)?