Question

The number of values of $x$ satisfying the equation $2^{\left({\log }_{5}16\right)\left({\log }_{4}x\right)+\left({\log }_{\sqrt[x]{2}}5\right)}+x^{\left({\log }_{5}4\right)+5}+x^5=0$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

$2^{{\log }_{5}16\cdot {\log }_{4}x+{\log }_x{\sqrt{2}}^5}+5^x+x^{{\log }_{3}4+5}+x^5=0$
$2^{2{\log }_{5}4\cdot {\log }_{4}x+x{\log }_{2}5}+5^x+x^{{\log }_{5}4}\cdot x^5+x^5=0$
$2^{2{\log }_{5}x}{2}^{x{\log }_{2}5}+5^x+x^{2{\log }_{5}2}\cdot x^5+x^5=0$
${\left(2^{{\log }_{5}x}\right)}^2\cdot 5^x+5^x+{\left(2^{{\log }_{5}x}\right)}^2\cdot x^5+x^5=0$
$5^x\left[{\left(2^{{\log }_{5}x}\right)}^2+1\right]+x^5\left[{\left(2^{{\log }_{5}x}\right)}^2+1\right]=0$
$\left(5^x+x^5\right)\left[{\left(2^{{\log }_{5}x}\right)}^2+1\right]=0$
$5^x+x^5=0{\left(2^{{\log }_{5}x}\right)}^2+1=0$
$\text{ This possible only when }x\text{ will be-ve }\text{ No solution }$
$\text{ while according to question }x\ge 2$
$\therefore \text{ number of values of }x=\text{ zero }$