Find-x-x-log-4-x-x-log-4-x-2-

Question Number 74244 by Maclaurin Stickker last updated on 20/Nov/19

F i n d x

x^{{l o g}_{4} \sqrt{x}} = x^{{l o g}_{4} x} - 2

Answered by MJS last updated on 20/Nov/19

t = x^{\log_{4} \sqrt{x}} = x^{\frac{\ln \sqrt{x}}{\ln 4}} = x^{\frac{\ln x}{4 \ln 2}}

\ln t = \frac{\ln x}{4 \ln 2} \ln x = \frac{{(\ln x)}^{2}}{4 \ln 2}

\Rightarrow x = e^{2 \sqrt{\ln 2 \ln t}}

x^{\log_{4} \sqrt{x}} = t

x^{\log_{4} x} = t^{2}

t = t^{2} - 2

t^{2} - t - 2 = 0

\Rightarrow t = - 1 \lor t = 2 but for x \in R we need t > 1

\Rightarrow t = 2 \Rightarrow x = 4

Answered by mr W last updated on 20/Nov/19

x > 0

l e t u = x^{\log_{4} \sqrt{x}} = x^{\frac{1}{2} \log_{4} x} = \sqrt{x^{\log_{4} x}} > 0

\Rightarrow u = u^{2} - 2

\Rightarrow u^{2} - u - 2 = 0

\Rightarrow (u - 2) (u + 1) = 0

\Rightarrow u = 2, - 1

w i t h u = 2 :

\sqrt{x^{\log_{4} x}} = 2

x^{\log_{4} x} = 4

x^{\frac{\ln x}{\ln 4}} = 4

\frac{\ln x}{\ln 4} = \frac{\ln 4}{\ln x}

{(\ln x)}^{2} = {(\ln 4)}^{2}

\ln x = \pm \ln 4 = \ln 4^{\pm 1}

\Rightarrow x = 4^{\pm 1} = 4 o r \frac{1}{4}

Commented by MJS last updated on 20/Nov/19

same path we share \dots

but the given equation also has the solution

x = \frac{1}{4}

how to get it ?

Commented by mr W last updated on 20/Nov/19

t w o s o l u t i o n s i n d e e d . s e e a b o v e .

Commented by Maclaurin Stickker last updated on 21/Nov/19

S i r, I a d d e d 2 b y t h e s i d e s a n d a p p l i e d

{l o g}_{2} i n b o t h s i d e s . I s t h i s c o r r e c t ?

I g o t t h e s a m e r e s u l t s

Commented by MJS last updated on 21/Nov/19

{doesn}^{'} t sound correct to me, please post it

Commented by Maclaurin Stickker last updated on 21/Nov/19

x^{{l o g}_{4} \sqrt{x}} + 2 = x^{{l o g}_{4} x}

{l o g}_{2} (x^{{l o g}_{4} \sqrt{x}}) + {l o g}_{2} 2 = {l o g}_{2} (x^{{l o g}_{4} x})

\dots .

\frac{1}{4} {l o g}_{2} x^{2} = \frac{1}{2} {l o g}_{2} x^{2} - 1

\dots .

{l o g}_{2} x^{2} = 4

{l o g}_{2} x = 2 a n d {l o g}_{2} x = - 2

x = 4 o r x = \frac{1}{4}

Commented by MJS last updated on 21/Nov/19

funny you get the solution \dots

but

\log_{b} (p + q) \neq \log_{b} p + \log_{b} q

b^{\log_{b} (p + q)} = p + a but b^{\log_{b} p + \log_{b} q} = p q

so your 2^{nd} line is not equivalent to the first

line

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