The-number-of-integral-solutions-of-the-equation-4log-x-2-x-2log-4x-x-2-3log-2x-x-3-is-

Question Number 22220 by Tinkutara last updated on 13/Oct/17

The number of integral solutions of the

equation {4 \log}_{x / 2} (\sqrt{x}) + {2 \log}_{4 x} (x^{2}) =

{3 \log}_{2 x} (x^{3}) is

Answered by ajfour last updated on 13/Oct/17

4 \log_{x / 2} [\sqrt{2} (\sqrt{x / 2})] + 2 \log_{4 x} [\frac{{(4 x)}^{2}}{16}]

= 3 \log_{2 x} [\frac{{(2 x)}^{3}}{8}]

\Rightarrow 4 \log_{x / 2} \sqrt{2} + 4 (\frac{1}{2}) + 2 \times 2 - 2 \log_{4 x} 16

= 3 \times 3 - 3 \log_{2 x} 8

\frac{4 \log_{2} \sqrt{2}}{\log_{2} x - 1} - \frac{2 \log_{2} 16}{\log_{2} x + 2} = 3 - \frac{3 \log_{2} 8}{\log_{2} x + 1}

l e t \log_{2} x = t, t h e n

\frac{2}{t - 1} - \frac{8}{t + 2} = 3 - \frac{9}{t + 1}

i f t \neq 1, - 2, - 1 t h e n

\frac{2}{t - 1} + \frac{9}{t + 1} = 3 + \frac{8}{t + 2} \Rightarrow

\frac{2 t + 2 + 9 t - 9}{t^{2} - 1} = \frac{3 t + 6 + 8}{t + 2}

(11 t - 7) (t + 2) = (t^{2} - 1) (3 t + 14)

11 t^{2} + 15 t - 14 = 3 t^{3} + 14 t^{2} - 3 t - 14

3 t^{3} + 3 t^{2} - 18 t = 0

\Rightarrow t (t^{2} + t^{2} - 6) = 0

o r t (t + 3) (t - 2) = 0

t = \log_{2} x = 0, 2, - 3

\Rightarrow x = 1, 4, \frac{1}{8}

h e n c e t w o i n t e g r a l s o l u t i o n s .

Commented by Tinkutara last updated on 14/Oct/17

Thank you very much Sir!