Respuesta :

kingkel
y=log^2(x+7) equals y=(7+x)*LogN^2 OR {{x=(-7.0*LogN^2.0+y)*LogN^(-2.0)}}
I'm assuming it's [tex]y = log_2(x + 7)[/tex]

[tex]x = log_2(y + 7)[/tex]
[tex]2^{x} = y + 7[/tex]
[tex]y = 2^{x} - 7[/tex]

If it actually is [tex]y = log^2(x + 7)[/tex],
the inverse is just [tex]y = e^{\sqrt{x}} - 7[/tex]

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