f-x-ln-1-ln-x-f-x-ln-f-x-Find-x-

Question Number 143293 by Huy last updated on 12/Jun/21

f (x) = l n (1 + l n (x)) .

f (x) = l n (f^{'} (x)) . F i n d x

Answered by Olaf_Thorendsen last updated on 12/Jun/21

f (x) = \ln (1 + \ln x)

f^{'} (x) = \frac{\frac{1}{x}}{1 + \ln x}

f (x) = \ln (f^{'} (x))

\ln (1 + \ln x) = \ln (\frac{1}{x (1 + \ln x)})

1 + \ln x = \frac{1}{x (1 + \ln x)}

x {(1 + \ln x)}^{2} = 1

x = 1 is an evident solution

Answered by mathmax by abdo last updated on 12/Jun/21

f (x) = \log (1 + logx) we have 1 + logx > 0 \Rightarrow logx > - 1 \Rightarrow x > e^{- 1}

f is defined on] e^{- 1}, + \infty [

f (x) = \log (f^{^{'}} (x)) \Rightarrow \log (1 + logx) = \log (\frac{1}{x (1 + logx)}) \Rightarrow

1 + logx = \frac{1}{x (1 + logx)} \Rightarrow x {(1 + logx)}^{2} = 1 \Rightarrow

x (1 + 2 logx + \log^{2} x) - 1 = 0 \Rightarrow {xlog}^{2} x + 2 xlogx + x - 1 = 0 = Ψ (x)

Ψ (x) = {xlog}^{2} x + 2 xlogx + x - 1 \Rightarrow

Ψ^{^{'}} (x) = \log^{2} x + 2 logx + 2 logx + 2 + 1 = \log^{2} x + 4 logx + 3

= u^{2} + 4 u + 3 (u = logx)

Δ^{^{'}} = 2^{2} - 3 = 1 \Rightarrow logx = - 2 + 1 or logx = - 2 - 1 \Rightarrow x = e^{- 1} or x = e^{- 3}

x e^{- 3} e^{- 1} + \infty

Ψ^{^{'}} 0 - 0 +

Ψ decr incr + \infty

Ψ is decreazing onI =] \frac{1}{e^{3}}, \frac{1}{e} [\Rightarrow \exists! x_{0} \in I / Ψ (x_{0}) = 0

Ψ is increazing onJ =] \frac{1}{e}, + \infty [\Rightarrow \exists! y_{0} \in J / Ψ (y_{0}) = 0

we see that Ψ (1) = 0 \Rightarrow y_{0} = 1

we can determine x_{0} by newton method \dots