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

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

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Answer 1 · 2017-12-09T15:00:59

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Explanation:

$x - ln (4 - x^{2}) = 0$ $x-ln(4-x^2)=0$ $\Leftrightarrow$ $<=>$

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

${ln e}^{x} = ln (4 - x^{2})$ $lne^x=ln(4-x^2)$

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

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

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

We have $f (x) = 0$ $f(x)=0$ $\Leftrightarrow$ $<=>$ $x^{2} - 4 + e^{x} = 0$ $x^2-4+e^x=0$

$\Leftrightarrow$ $<=>$ $(x - 2) (x + 2) + e^{x} = 0$ $(x-2)(x+2)+e^x=0$

From the definition of $ln$ $ln$ we know this
$y = ln x$ $y=lnx$ $\Leftrightarrow x = e^{y}$ $<=>x=e^y$

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

now that makes the equation transform to

$(ln y - 2) (ln y + 2) + y = 0$ $(lny-2)(lny+2)+y=0$

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

So, either $(ln y - 2) (ln y + 2) = 0$ $(lny-2)(lny+2)=0$ and $y = 0$ $y=0$

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

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

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

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

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

$g (y) = 0 \Leftrightarrow y = e^{2}$ $g(y)=0 <=> y=e^2$ (unique root)

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A Graphical interpritation:

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

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

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

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