Question

How many real roots does the following equation have? $e^x-e^{-x}+1=0$

• A. $0$
• B. $1$
• C. $2$
• D. $4$
• E. an infinite number

Answer 1

Answer: (B)

$1$

Equation is:

$e^x-e^{-x}+1=0$

$e^x-\frac{1}{e^x}+1=0$

$\frac{e^{2x}-1+e^x}{e^x}=0$

$e^{2x}+e^x-1=0$

Let's put

$e^x=y$

Then, we have

$y^2+y-1=0$

By using the quadratic equation

$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

i.e.,

$y=\frac{-1\pm \sqrt{1+4}}{2}$

$y_1=\frac{-1+\sqrt{5}}{2}$

and

$y_2=\frac{-1-\sqrt{5}}{2}$

Now find $x$

we have

$e^x=y$

$\ln y=x$

Thus

$x_1=ln\left[\left(\frac{-1+\sqrt{5}}{2}\right)\right]$

Exact form

$x_1=ln\left(2\right)$

Decimal form

$x_1=0.693$

and

$x_2=\ln \left[\left(\frac{-1-\sqrt{5}}{2}\right)\right]$

we know that the natural logarithm of a negative number is undefined.

i.e.,

$x_2=\alpha$

so, we conclude that for $e^x-e^{-x}+1=0$ we have only one real root.

Thus option $\left(B\right)$ is correct.