Question

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

Answer 1

See below

Explanation:

`x-ln(4-x^2)=0` `<=>`

`x=ln(4-x^2)`

`lne^x=ln(4-x^2)`

`e^x=4-x^2`

`x^2+e^x-4=0`

`f(x)=x^2+e^x-4` , `x` `in` `R`
So we need to study `f`

We have `f(x)=0` `<=>` `x^2-4+e^x=0`

`<=>` `(x-2)(x+2)+e^x=0`

From the definition of `ln` we know this
`y=lnx` `<=>x=e^y`

So if i set `e^x=y` i have `x=lny`

now that makes the equation transform to

`(lny-2)(lny+2)+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 `(lny-2)(lny+2)=0` and `y=0`

which means `lny=2` or `lny=-2` and `y=0`
`lny=-2` `->` Impossible!
So we have `lny=2 <=> y=e^2` and `y=0`

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

`<=>` `ln^2(y)-4=-y`
However, `y>0` , `-y<0`

So we need `ln^2(y)-4>0`

`g(y)=ln^2(y)-4, y>0`

`g(y)=0 <=> y=e^2` (unique root)

A Graphical interpritation:

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

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