Question-68234

Question Number 68234 by Mikael last updated on 07/Sep/19

Commented by kaivan.ahmadi last updated on 07/Sep/19

3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}} = 4^{x} + 2^{2 x - 1} \Rightarrow

3^{x - \frac{1}{2}} (3 + 1) = 2^{2 x - 1} (2 + 1) \Rightarrow

\frac{2^{2 x - 1}}{3^{x - \frac{1}{2}}} = \frac{4}{3} \Rightarrow \frac{2^{2 x - 1}}{4} = \frac{3^{x - \frac{1}{2}}}{3} \Rightarrow 2^{2 x - 3} = 3^{x + \frac{1}{2}} \Rightarrow

{\begin{cases} 2 x - 3 = 0 \Rightarrow x = \frac{3}{2} \\ x + \frac{1}{2} = 0 \Rightarrow x = - \frac{1}{2} \end{cases}

x = \frac{3}{2} i s a n s w e r

Commented by Mikael last updated on 07/Sep/19

t h a n k s

Commented by Prithwish sen last updated on 08/Sep/19

2^{2 x} = a 3^{x} = b

a - \frac{b}{\sqrt{3}} = \sqrt{3} b - \frac{a}{2}

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

\frac{a}{b} = \frac{2^{3}}{3^{\frac{3}{2}}}

∴ 2 x = 3 \Rightarrow x = \frac{3}{2}

Commented by Rasheed.Sindhi last updated on 09/Sep/19

Sir kaivan ahmadi I {don}^{'} t understand

2^{2 x - 3} = 3^{x + \frac{1}{2}} \overset{?}{\Rightarrow} {\begin{cases} 2 x - 3 = 0 \Rightarrow x = \frac{3}{2} \\ x + \frac{1}{2} = 0 \Rightarrow x = - \frac{1}{2} \end{cases}

Commented by kaivan.ahmadi last updated on 13/Sep/19

s i n c e 2^{0} = 3^{0}

o t h e r w i s e t h i s e q u a l i t y i s n o t t r u e .

Answered by Rasheed.Sindhi last updated on 09/Sep/19

4^{x} + 4^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}

4^{x - \frac{1}{2}} (4^{\frac{1}{2}} + 1) = 3^{x - \frac{1}{2}} (3 + 1)

4^{x - \frac{1}{2} - 1} = 3^{x - \frac{1}{2} - 1}

\frac{4^{x - \frac{3}{2}}}{3^{x - \frac{3}{2}}} = 1

{(\frac{4}{3})}^{x - \frac{3}{2}} = {(\frac{4}{3})}^{0}

x - \frac{3}{2} = 0

x = \frac{3}{2}