સમીકરણ {9^x} - {2^{x + {1 over 2}}} = {2^{x + | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) ${9^x} - {2^{x + (1/2)}} = {2^{x + (3/2)}} - {3^{2x - 1}}$

 ==> ${3^{2x}} + {1 \over 3}{.3^{2x}} = {2.2^{x + {1 \over 2}}} + {2^{x + {1

Vedclass · Accepted Answer

${\log _{\left( {9/2} \right)}}(9/\sqrt 8 )$

Vedclass · Answer

(b) ${9^x} - {2^{x + (1/2)}} = {2^{x + (3/2)}} - {3^{2x - 1}}$

 ==> ${3^{2x}} + {1 \over 3}{.3^{2x}} = {2.2^{x + {1 \over 2}}} + {2^{x + {1 \over 2} - 2}}$

==> $4\,.\,{3^{2x - 1}} = 3.\,{2^{x + {1 \over 2}}}$ ==> ${3^{2x - 2}} = {2^{x + {1 \over 2} - 2}}$

==> ${3^{2x - 2}} = {2^{x - {3 \over 2}}}$ ==> ${\left( {{9 \over 2}} ight)^{x - 1}} = {2^{ - 1/2}}$

==> $(x - 1)\,{\log _{9/2}}9/2 = - {1 \over 2}{\log _{9/2}}2$

==> $x - 1 = - {1 \over 2}{\log _{9/2}}2$

==> $x = 1 - {\log _{9/2}}\sqrt 2 = {\log _{9/2}}9/2 - {\log _{9/2}}\sqrt 2 $

==> $x = {\log _{9/2}}(9/2\sqrt 2 )$;

$	herefore x = {\log _{9/2}}(9/\sqrt 8 )$.

સમીકરણ ${9^x} - {2^{x + {1 \over 2}}} = {2^{x + {3 \over 2}}} - {3^{2x - 1}}$ નો ઉકેલ મેળવો.

Similar Questions

$\sqrt {(3 + \sqrt 5 )} = . .$ .

જો $x = 3 - \sqrt {5,} $ તો ${{\sqrt x } \over {\sqrt 2 + \sqrt {(3x - 2)} }} = $

$\sqrt {[12\sqrt 5 + 2\sqrt {(55)} ]} $ નું વર્ગમૂળ મેળવો.

$2\sqrt 3 - \sqrt 7 $ નો સંમેય કારક અવયવ મેળવો.

${{{{2.3}^{n + 1}} + {{7.3}^{n - 1}}} \over {{3^{n + 2}} - 2{{(1/3)}^{l - n}}}} = $