How do you solve # log x = x^2 - 2#?

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Answer 1 · 2016-05-27T07:17:55

#-i sqrt(W(-2/e^4)/2) and -i sqrt(W(-1,-2/e^4)/2) #

This is where it gets messy

There is not elementary way to solve this problem

Probably your best shot at solving this problem is graphing it.And says a a lot about this problem

Fortunately I do have a solution

But first lets see why there is no scope to solve this using logarithmic properties

#log x = x^2 - 2#

I am going to raise 10 to the power of each side

#10^logx = 10^(x^2 -2)#

# x= 10^(x^2 -2)#

Now if you want to try solving this you would probably end up going in cirles

So here is where the real solution start

Have you heard about the Lambert W function or product log

Here is a brief explanation

enter image source here

You can read more about this
HERE

So lets Begin

log

#log⁡(x)=x^2−2#

Another thing In your case I am assuming logarithm to be Natural logarithm

#lnx =x^2 -2#

#x = e^(x^2-2)#

#x*e^(2-x^2) =1#

Now apply the the Lambert W function on both the sides

Since this is very big calculation I used wolfram alpha

So the 2 roots are

#-i sqrt(W(-2/e^4)/2) and -i sqrt(W(-1,-2/e^4)/2) #