સમીકરણ {4.9^{x - 1}} = 3sqrt {({2^{2x + 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}$.

સમીકરણ ${4.9^{x - 1}} = 3\sqrt {({2^{2x + 1}})} $ નો ઉકેલ મેળવો.

Similar Questions

જો ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ તો $x = .. . .$

$\sqrt {[12 - \sqrt {(68 + 48\sqrt 2 )} ]} = $

જો ${a^x} = {(x + y + z)^y},{a^y} = {(x + y + z)^z}$, ${a^z} = {(x + y + z)^x},$ તો

જો $a = \sqrt {(21)} - \sqrt {(20)} $ અને $b = \sqrt {(18)} - \sqrt {(17),} $ તો

${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $