log-4-8-x-1-log-4-x-solve-for-x-

Question Number 176748 by MASANJAJJ last updated on 26/Sep/22

{l o g}_{4} \sqrt{8 - x} = 1 - {l o g}_{4} x

s o l v e f o r x

Answered by mr W last updated on 26/Sep/22

0 < x < 8

\log_{4} \sqrt{8 - x} = \log_{4} \frac{4}{x}

\sqrt{8 - x} = \frac{4}{x}

8 - x = \frac{16}{x^{2}}

\frac{1}{x^{3}} - \frac{1}{2 x} + \frac{1}{16} = 0

\frac{1}{x} = \frac{2}{\sqrt{6}} \sin (\frac{1}{3} \sin^{- 1} \frac{3 \sqrt{6}}{16} + \frac{2 k π}{3}), k = 0, 1 (2 r e j e c t e d)

\Rightarrow x = \frac{\sqrt{6}}{2 \sin (\frac{1}{3} \sin^{- 1} \frac{3 \sqrt{6}}{16} + \frac{2 k π}{3})}, k = 0, 1

Commented by Tawa11 last updated on 27/Sep/22

Great sir

Answered by Rasheed.Sindhi last updated on 26/Sep/22

{l o g}_{4} \sqrt{8 - x} = 1 - {l o g}_{4} x

{l o g}_{4} \sqrt{8 - x} + {l o g}_{4} x = 1

\log_{4} x \sqrt{8 - x} = \log_{4} 4

x \sqrt{8 - x} = 4

x^{2} (8 - x) = 16

x^{3} - 8 x^{2} + 16 = 0

x \approx 1.73, 1.58, - 1.31

Commented by Rasheed.Sindhi last updated on 26/Sep/22

S i r, t h i s i s w h a t t h e c a l c u l a t o r

c a l c u l a t e s . A c t u a l l y I f e e d x^{3} - 8 x^{2} + 16 = 0

t o t h e c a l c u l a t o r i n s t e a d o f t h e o r i g i n a l

e q u a t i o n!

Commented by mr W last updated on 26/Sep/22

x < 0 {d o e s n}^{'} t f i t t h e o r i g i n a l e q u a t i o n .

y o u m e a n t x \approx 7.73 ?

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