સમીકરણ x+1-2 log _{2}left(3+2^{x}right)+2 log _{4}left(10-2^{-x}right)=0 ના ઉક?

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4-2.Quadratic Equations and Inequations

hard

સમીકરણ $x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$ ના ઉકેલનો સરવાળો મેળવો.

$\log _{2} 14$

$\log _{2} 11$

$\log _{2} 12$

$\log _{2} 13$

(JEE MAIN-2021)

$x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$

$\log _{2}\left(2^{x+1}\right)-\log _{2}\left(3+2^{x}\right)^{2}+\log _{2}\left(10-2^{-x}\right)=0$

$\log _{2}\left(\frac{2^{x+1} \cdot\left(10-2^{-x}\right)}{\left(3+2^{x}\right)^{2}}\right)=0$

$\frac{2\left(10.2^{x}-1\right)}{\left(3+2^{x}\right)^{2}}=1$

$\Rightarrow 20.2^{x}-2=9+2^{2 x}+6.2^{x}$

$\therefore\left(2^{x}\right)^{2}-14\left(2^{x}\right)+11=0$

Roots are $2^{x_{1}} \& 2^{x_{2}}$

$\therefore 2^{x_{1}} \cdot 2^{x_{2}}=11$

$x_{1}+x_{2}=\log _{2}(11)$

Standard 11

Mathematics

hard

(JEE MAIN-2023)

medium

medium

medium

(IIT-2004)

hard

(JEE MAIN-2021)