How do you solve x-ln(4-x^2)=0?

Dec 9, 2017

See below

Explanation:

$x - \ln \left(4 - {x}^{2}\right) = 0$ $\iff$

$x = \ln \left(4 - {x}^{2}\right)$

$\ln {e}^{x} = \ln \left(4 - {x}^{2}\right)$

${e}^{x} = 4 - {x}^{2}$

${x}^{2} + {e}^{x} - 4 = 0$

$f \left(x\right) = {x}^{2} + {e}^{x} - 4$ , $x$$\in$$R$
So we need to study $f$

We have $f \left(x\right) = 0$ $\iff$ ${x}^{2} - 4 + {e}^{x} = 0$

$\iff$ $\left(x - 2\right) \left(x + 2\right) + {e}^{x} = 0$

From the definition of $\ln$ we know this
$y = \ln x$ $\iff x = {e}^{y}$

So if i set ${e}^{x} = y$ i have $x = \ln y$

now that makes the equation transform to

$\left(\ln y - 2\right) \left(\ln y + 2\right) + y = 0$

• We have the sum of $a + b = 0$
In order for this to be true we need either $a = 0$ and $b = 0$
or the numbers to have opposite sign between them

So, either $\left(\ln y - 2\right) \left(\ln y + 2\right) = 0$ and $y = 0$

which means $\ln y = 2$ or $\ln y = - 2$ and $y = 0$
$\ln y = - 2$ $\to$ Impossible!
So we have $\ln y = 2 \iff y = {e}^{2}$ and $y = 0$

If the numbers are opposite then that means $\left(\ln y - 2\right) \left(\ln y + 2\right) + y = 0$

$\iff$ ${\ln}^{2} \left(y\right) - 4 = - y$
However, $y > 0$ , $- y < 0$

So we need ${\ln}^{2} \left(y\right) - 4 > 0$

$g \left(y\right) = {\ln}^{2} \left(y\right) - 4 , y > 0$

$g \left(y\right) = 0 \iff y = {e}^{2}$ (unique root)

A Graphical interpritation:

Graph of $y = x - \ln \left(4 - {x}^{2}\right)$ where roots are the solutions

The graphs, $y = x \mathmr{and} y = \ln \left(4 - {x}^{2}\right)$ where the intisections are the solutions