Solve-x-R-x2-1-x-1-x-2-x-4-Source-L-Panaitopol-

Question Number 173485 by mnjuly1970 last updated on 12/Jul/22

S o l v e (x \in R)

x 2^{\frac{1}{x}} + \frac{1}{x} 2^{x} = 4

- S o u r c e : L . P a n a i t o p o l

Answered by aleks041103 last updated on 12/Jul/22

o b v x > 0 f o r x 2^{1 / x} + 2^{x} / x = 4

f (x) = x 2^{1 / x}

\Rightarrow g (x) = f (x) + f (1 / x)

\Rightarrow g (1 / x) = f (1 / x) + f (x)

\Rightarrow g (x) = g (1 / x) = 4

\Rightarrow t h e r e f o r e i t i s e n o u g h t o s e a r c h

f o r x > 1 a n d f o r \forall x_{i} > 1, s . t . g (x_{i}) = 4 {w e}^{'} l l

h a v e {\bar{x}}_{i} = \frac{1}{x_{i}}, s . t . g ({\bar{x}}_{i}) = 4 t o o .

2^{x} = e^{l n (2) x} = \underset{i = 0}{\sum^{\infty}} \frac{{(l n (2) x)}^{i}}{i!}

\Rightarrow f (x) = \underset{i = 0}{\sum^{\infty}} \frac{{(l n (2))}^{i}}{i!} \frac{1}{x^{i - 1}}

\Rightarrow f (1 / x) = \underset{i = 0}{\sum^{\infty}} \frac{{(l n (2))}^{i}}{i!} x^{i - 1}

\Rightarrow g (x) = \underset{i = 0}{\sum^{\infty}} \frac{{(l n (2))}^{i}}{i!} (x^{i - 1} + \frac{1}{x^{i - 1}})

g (x) = \underset{i = 0}{\sum^{\infty}} \frac{{(l n (2))}^{i}}{i!} s_{i - 1} (x)

s_{i} (x) = x^{i} + x^{- i}

s_{i}^{'} (x) = {i x}^{i - 1} - {i x}^{- i - 1} = \frac{i}{x} (x^{i} - x^{- i})

f o r x > 1 :

i > 0 : x^{i} > 1 \Rightarrow x^{- i} < 1 \Rightarrow x^{i} - x^{- i} > 0

\Rightarrow s_{i > 0}^{'} (x > 1) > 0

i < 0 : x^{i < 1} \Rightarrow x^{- i} > 1 \Rightarrow x^{i} - x^{- i} < 0

\Rightarrow i (x^{i} - x^{- i}) > 0

\Rightarrow s_{i < 0}^{'} (x > 1) > 0

i = 0 :

s_{0}^{'} = 0

\Rightarrow g^{'} (x > 1) = \underset{i = 0, i \neq 1}{\sum^{\infty}} \frac{{(l n (2))}^{i}}{i!} s_{i - 1}^{'} (x > 1)

s i n c e s_{i - 1}^{'} (x > 1) > 0 \Rightarrow g^{'} (x > 1) > 0

\Rightarrow g (x) i s s t r i c t l y i n c r e a s i n g f o r x > 1

\Rightarrow f o r g (x) = 4 e x i s t s a t m o s t 1 s o l n . x ⩾ 1

A n d g (1) = 4

\Rightarrow T h e o n l y s o l u t i o n t o t h e p r o b l e m i s

x = 1

Commented by Tawa11 last updated on 13/Jul/22

Great sir

Answered by dragan91 last updated on 12/Jul/22

2^{\log_{2} x + \frac{1}{x}} + 2^{\log_{2} \frac{1}{x} + x} = 4

2^{\log_{2} x + \frac{1}{x}} + 2^{- \log_{2} x + x} = 4

2^{\log_{2} x + \frac{1}{x}} + 2^{- \log_{2} x + x} \overset{AM - GM}{⩾} 2 \sqrt{2^{x + \frac{1}{x}}}

2^{x + \frac{1}{x}} ⩽ 2^{2} ∣ \log_{2}

x + \frac{1}{x} ⩽ 2

since x + \frac{1}{x} \overset{AM - GM}{⩾} 2 \Rightarrow equation has solution

only for x = 1

Commented by Tawa11 last updated on 13/Jul/22

Great sir

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