Solve-for-real-numbers-3-x-x-3-1-1-x-12-

Question Number 163452 by HongKing last updated on 07/Jan/22

Solve for real numbers :

3^{x \sqrt{x}} + 3^{1 + \frac{1}{\sqrt{x}}} = 12

Commented by mr W last updated on 07/Jan/22

i {d o n}^{'} t t h i n k s u c h e q u a t i o n s h a v e

m u c h t e c h n i q u e v a l u e s . b a s i c a l l y

w e m u s t “ {s e e}^{″} o r g u e s s t h e s o l u t i o n s .

h e r e f o r e x a m p l e : w e g u e s s 3^{2} + 3 = 12

t h e n s e e x = 1 i s o k .

i s a y t h i s h a s n o t e c h n i q u e v a l u e,

b u t r a t h e r s t u p i d, b e c a u s e w e a r e d e a d

i f t h e e q u a t i o n i s

3^{x \sqrt{x}} + 3^{1 + \frac{1}{\sqrt{x}}} = 11 .

i h a v e n e v e r u n d e r s t o o d w h a t t h e

c r e a t o r s (i {d o n}^{'} t m e a n H o n g k i n g s i r)

o f s u c h q u e s t i o n s w a n t t o t e s t f r o m

t h e s t u d e n t s .

Answered by mathlove last updated on 08/Jan/22

s o l v e \frac{1}{\sqrt{x}} = t \Rightarrow \frac{1}{x} = t^{2} \Rightarrow x = \frac{1}{t^{2}}

3^{\frac{1}{t^{2}} \times \frac{1}{t}} + 3^{1} \times 3^{t} = 12

3^{\frac{1}{t^{3}}} + 3^{1} \times 3^{t} = 12

3^{t} (3^{\frac{1}{t^{3}} - t} + 3) = (3 \times 4)

3^{t} = 3^{1} \Rightarrow t = 1

3^{\frac{1}{t^{3}} - t} = 1 \Rightarrow 3^{\frac{1}{t^{3}} - t} = 3^{0}

\frac{1}{t^{3}} - t = 0 \Rightarrow 1 - t^{4} = 0 \Rightarrow t^{4} = 1 \Rightarrow t = 1

h o w x = \frac{1}{t^{2}} \Rightarrow x = 1

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