Solution of the equation {9^x} - {2^{x + {1 o | 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 \over

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 )$.

Solution of the equation ${9^x} - {2^{x + {1 \over 2}}} = {2^{x + {3 \over 2}}} - {3^{2x - 1}}$

Similar Questions

${4 \over {1 + \sqrt 2 - \sqrt 3 }} = $

The greatest number among $\root 3 \of 9 ,\root 4 \of {11} ,\root 6 \of {17} $ is

If ${a^x} = {(x + y + z)^y},{a^y} = {(x + y + z)^z}$, ${a^z} = {(x + y + z)^x},$ then

If $x = \sqrt 7 + \sqrt 3 $ and $xy = 4,$then ${x^4} + {y^4}=$

The cube root of $9\sqrt 3 + 11\sqrt 2 $ is