Question

The number of solutions of the equation $2x-\ln (2e^x-3)=c,$ where $c<0$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$2x-c=\ln (2e^x-3)$
$\Rightarrow e^{(2x-c)}=2e^x-3$
$\Rightarrow \frac{e^{2x}}{e^c}+3=2e^x$
$\Rightarrow \frac{e^x}{e^c}+\frac{3}{e^x}=2$

By AM-GM inequality,
$\frac{e^x}{e^c}+\frac{3}{e^x}\ge 2\sqrt{\frac{3}{e^c}}\Rightarrow 2\sqrt{\frac{3}{e^c}}\le \frac{e^x}{e^c}+\frac{3}{e^x}\Rightarrow 2\sqrt{\frac{3}{e^c}}\le 2\Rightarrow \sqrt{\frac{3}{e^c}}\le 1\Rightarrow e^c\ge \mathrm{3.....}\left(1\right)$

As $c<0,\Rightarrow e^c<\mathrm{1...}\left(2\right)$

equation $\left(1\right)$ and $\left(2\right)$ contradicts eachother. Hence, the equation has no solution