Question

If ${\log }_{(e+\pi )}{\log }_2(\sqrt{4x+1}+\sqrt{4x})=0,xϵR$ then $64x$ is equal to:

• A. $1$
• B. $3$
• C. $9$
• D. $12$

Answer 1

The correct option is C $9$

Given ${\log }_{(e+\pi )}{\log }_2(\sqrt{4x+1}+\sqrt{4x})=0$

$\Rightarrow {\log }_2(\sqrt{4x+1}+\sqrt{4x})=1$

$\Rightarrow \sqrt{4x+1}+\sqrt{4x}=2^1=2$

Squaring both sides, we get

$4x+1+4x+2\sqrt{4x(4x+1)}=4$

$\Rightarrow 8x-3+2\sqrt{4x(4x+1)}=0$

$\Rightarrow 2\sqrt{4x(4x+1)}=3-8x$

Squaring both sides:

$4\left(4x\right(4x+1\left)\right)=9-48x+64x^2$

$\Rightarrow 4(16x^2+4x)=64x^2-48x+9$

$\Rightarrow 64x^2+16x=64x^2-48x+9$

$\Rightarrow 64x=9$