Solution of the equation {4.9^{x - 1}} | vedclass

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

Question

(c) $4.\,{9^{x - 1}} = 3\,.\,\sqrt {({2^{2x + 1}})} $$ \Rightarrow $${3^{2x - 2 - 1}} = {2^{{{2x + 1} \over 2} - 2}}$

$ \Rightarrow $ ${3^{2x - 3}}

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(c) $4.\,{9^{x - 1}} = 3\,.\,\sqrt {({2^{2x + 1}})} $$ \Rightarrow $${3^{2x - 2 - 1}} = {2^{{{2x + 1} \over 2} - 2}}$

$ \Rightarrow $ ${3^{2x - 3}} = {2^{{{2x - 3} \over 2}}}$ $ \Rightarrow $ ${2^{{{2x - 3} \over 2}}} = {\left( {{3^{{{2x - 3} \over 2}}}} ight)^2}$
 $ \Rightarrow $$2x - 3 = 0$,

$	herefore x = {3 \over 2}$.

Solution of the equation ${4.9^{x - 1}} = 3\sqrt {({2^{2x + 1}})} $ has the solution

Similar Questions

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

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

$\sqrt {(3 + \sqrt 5 )} - \sqrt {(2 + \sqrt 3 )} = $

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

${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $