Given that f(x) is 4^8_+1 what is f^-1_(x)?

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  • 8 months ago
    Favourite answer

    If you meant

    f(x) = 4^(8x+1), then

    8x+1 = log_4(f(x)), and

    x = [log_4(f(x)) - 1]/8, so

    f^(-1) (x) = [log_4(x) - 1]/8.

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  • Anonymous
    8 months ago

    f(x) = 4⁸ + 1

    f⁻¹(x) = ?f(x) = yx = f⁻¹(y)y = 4⁸ + 1Here there is ‘x’ termLet f(x) = 4x + 1f⁻¹(x) = ?y = 4x + 1y-1 = 4xlog(y-1) = xlog4x = log(y-1) / log4x = log(y-1-4)x = log(y-5)f⁻¹(x) = log(x-5).

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  • Ash
    Lv 7
    8 months ago

    There is no x in f(x). Can you please check your problem ?

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