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Question Number 180301 by Noorzai last updated on 10/Nov/22

Commented by MJS_new last updated on 10/Nov/22
$this has the shape a^x +(1/a^x )=b^x which we can solve for b: b=((a^x +(1/a^x )))^(1/x) and for a: a^x +(1/a^x )=2cosh (xln a) =b^x ⇒ a=e^((cosh^(−1) (b^x /2))/x) but we cannot solve for x. we can { ((always approximate)),((sometimes see an obvious solution)) :}$
$this has the shape$ $a^{x} + \frac{1}{a^{x}} = b^{x}$ $which we can solve for b :$ $b = \sqrt[x]{a^{x} + \frac{1}{a^{x}}}$ $and for a :$ $a^{x} + \frac{1}{a^{x}} = 2 \cosh (x \ln a) = b^{x}$ $\Rightarrow a = e^{\frac{\cosh^{- 1} \frac{b^{x}}{2}}{x}}$ $but we cannot solve for x .$ $we can {\begin{cases} always approximate \\ sometimes see an obvious solution \end{cases}$
	Answered by MJS_new last updated on 10/Nov/22

	$obviously x = 2 :$ ${(\sqrt{2 + \sqrt{3}})}^{2} + {(\sqrt{2 - \sqrt{3}})}^{2} = 2 + \sqrt{3} + 2 - \sqrt{3} = 4$
	Commented by peter frank last updated on 10/Nov/22

	$how ?$
	Commented by MJS_new last updated on 10/Nov/22

	${(\sqrt{t})}^{2} = t$ $what else do you need to know ?$
	Answered by Rasheed.Sindhi last updated on 10/Nov/22

	${(\sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}})}^{x} + 1 = {(\frac{2}{\sqrt{2 - \sqrt{3}}})}^{x}$ ${(\sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}}})}^{x} + 1 = {(\frac{2}{\sqrt{2 - \sqrt{3}}} \times \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2 + \sqrt{3}}})}^{x}$ ${(2 + \sqrt{3})}^{x} + 1 = {(2 \sqrt{2 + \sqrt{3}})}^{x}$ ${(2 + \sqrt{3})}^{x} - {(2 \sqrt{2 + \sqrt{3}})}^{x} + 1 = 0$ ${(2 + \sqrt{3})}^{x} - {(4 (2 + \sqrt{3}))}^{x / 2} + 1 = 0$ ${(2 + \sqrt{3})}^{x} - 4^{x / 2} {(2 + \sqrt{3})}^{x / 2} + 1 = 0$ ${(2 + \sqrt{3})}^{x / 2} = y$ $y^{2} - 4^{x / 2} y + 1 = 0$ $y = \frac{4 \pm \sqrt{4^{x} - 4}}{2} = {(2 + \sqrt{3})}^{x / 2}$ $2 \pm \sqrt{4^{x - 1} - 1} = {(2 + \sqrt{3})}^{x / 2}$ $L H S c o n s i s t o f t w o t e r m s :$ $n o n - r a d i c a l & r a d i c a l$ $a n d n o n - r a d i c a l = 2$ $T h i s i s p o s s i b l e w h e n$ $\frac{x}{2} = 1 \Rightarrow x = 2$ $a n d t h i s s a t i s f i e d a l s o r a d i c a l t e r m s$ $2 \pm \sqrt{4^{2 - 1} - 1} = {(2 + \sqrt{3})}^{2 / 2}$ $2 + \sqrt{3} = {(2 + \sqrt{3})}^{2 / 2}$ $[∵ 2 - \sqrt{3} = {(2 + \sqrt{3})}^{2 / 2} {d o e s n}^{'} t s a t i s y]$ $∴ x = 2$
	Answered by mr W last updated on 10/Nov/22

	$w e k n o w y = a^{x} i s s t r i c t l y i n c r e a s i n g$ $f o r a > 1 .$ ${(2 + \sqrt{3})}^{\frac{x}{2}} + {(2 - \sqrt{3})}^{\frac{x}{2}} = 4^{\frac{x}{2}}$ ${i t}^{'} s o b v i o u s t h a t x = 2 i s a s o l u t i o n .$ $s i n c e y = {(2 + \sqrt{3})}^{\frac{x}{2}} + {(2 - \sqrt{3})}^{\frac{x}{2}} - 4^{\frac{x}{2}}$ $i s s t r i c t l y i n c r e a s i n g o r s t r i c t l y$ $d e c r e a s i n g, i t h a s o n e a n d o n l y o n e$ $z e r o, s o x = 2 i s t h e o n l y s o l u t i o n .$
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