solvd-for-x-2-3-x-2-3-x-4-

Question Number 25317 by ibraheem160 last updated on 08/Dec/17

s o l v d f o r x : {(\sqrt{2 + \sqrt{3}})}^{x} + {(\sqrt{2 - \sqrt{3}})}^{x} = 4

Answered by ajfour last updated on 08/Dec/17

l e t u = {(\sqrt{2 + \sqrt{3}})}^{x}

\Rightarrow \ln u = x \ln \sqrt{2 + \sqrt{3}}

o r \ln u = \frac{x}{2} \ln (2 + \sqrt{3})

v = {(\sqrt{2 + \sqrt{3}})}^{x}

\Rightarrow \ln v = \frac{x}{2} \ln (2 - \sqrt{3})

= - \frac{x}{2} \ln (2 + \sqrt{3})

\Rightarrow \ln u v = 0

o r u v = 1 a n d u + v = 4 (g i v e n)

s o (u - v) = \sqrt{16 - 4} = 2 \sqrt{3}

h e n c e u = 2 + \sqrt{3} = {(\sqrt{2 + \sqrt{3}})}^{x}

\Rightarrow x = 2 .

Answered by Rasheed.Sindhi last updated on 08/Dec/17

Without using logarthm

Let {(\sqrt{2 + \sqrt{3}})}^{x} = a

a = {(\sqrt{\frac{(2 + \sqrt{3}) (2 - \sqrt{3})}{2 - \sqrt{3}}})}^{x}

a = {(\sqrt{\frac{1}{2 - \sqrt{3}}})}^{x}

a = \frac{1}{{(\sqrt{2 - \sqrt{3}})}^{x}}

{(\sqrt{2 - \sqrt{3}})}^{x} = \frac{1}{a}

a + \frac{1}{a} = 4

a^{2} - 4 a + 1 = 0

a = \frac{4 \pm \sqrt{16 - 4}}{2}

a = \frac{4 \pm 2 \sqrt{3}}{2} = 2 \pm \sqrt{3}

{(\sqrt{2 + \sqrt{3}})}^{x} = 2 + \sqrt{3}, 2 - \sqrt{3}

{(2 + \sqrt{3})}^{x / 2} = {(2 + \sqrt{3})}^{1}

x / 2 = 1 \Rightarrow x = 2

{(\sqrt{2 + \sqrt{3}})}^{x} = 2 - \sqrt{3}

{(2 + \sqrt{3})}^{x / 2} = {(2 + \sqrt{3})}^{- 1}

x / 2 = - 1 \Rightarrow x = - 2

x = \pm 2