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

$\root 4 \of {(17 + 12\sqrt 2 )} = $

Solution of the equation $\sqrt {(x + 10)} + \sqrt {(x - 2)} = 6$ are

$\sqrt {(3 + \sqrt 5 )} $ is equal to

If ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ then $x =$

${{\sqrt {(5/2)} + \sqrt {(7 - 3\sqrt 5 )} } \over {\sqrt {(7/2)} + \sqrt {(16 - 5\sqrt 7 )} }}=$