x-2-16-x-find-x-

Question Number 138627 by KwesiDerek last updated on 15/Apr/21

\boldsymbol x^{2} = 16^{\boldsymbol x}

\boldsymbol find \boldsymbol x

Commented by soudo last updated on 15/Apr/21

nice

Answered by mr W last updated on 15/Apr/21

x = \pm 16^{\frac{x}{2}} = \pm 4^{x}

x = \pm e^{x \ln 4}

(- x \ln 4) e^{- x \ln 4} = \pm \ln 4

- x \ln 4 = W (\pm \ln 4)

\Rightarrow x = - \frac{W (\pm \ln 4)}{\ln 4} = {\begin{cases} - \frac{W (\ln 4)}{\ln 4} = - \frac{1}{2} \\ - \frac{W (- \ln 4)}{\ln 4} \Rightarrow n o r e a l \end{cases}

Commented by KwesiDerek last updated on 15/Apr/21

thank you sir

Answered by MJS_new last updated on 16/Apr/21

x^{2} - 16^{x} = 0

(x - 4^{x}) (x + 4^{x}) = 0

x - 4^{x} = 0 \lor x + 4^{x} = 0

f_{1} (x) = x - 4^{x}

f_{1}^{'} (x) = 1 - 4^{x} \ln 4 = 0 \Rightarrow x_{0} = - \frac{\ln \ln 4}{\ln 4}

f_{1}^{″} (x) < 0 \forall x \in R \Rightarrow \max at x_{0} with f_{1} (x_{0}) < 0

\Rightarrow no solution for f_{1} (x) = 0

f_{2} (x) = x + 4^{x}

f_{2}^{'} (x) = 1 + 4^{x} \ln 4 > 0 \forall x \in R \land - \infty < f_{2} (x) < + \infty \Rightarrow

\Rightarrow exactly one solution for f_{2} (x) = 0

x = - 4^{x} \Leftrightarrow x = - 2^{2 x} \Rightarrow obviously x = - \frac{1}{2}

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