Solving for y: e^(y) = y^(2) - 2

[flyingfishcattle]

Homework Statement

Hi
I am having trouble solving for y for the following equation:
e^(y) = y^(2) - 2

Homework Equations

ln(x)^(y) = y*ln(x)
ln(e^(y)) = y
ln(2) = 1

The Attempt at a Solution

My attempt:
e^(y) = y^(2) -2
ln(e^(y)) = ln(y^(2)) - ln(2)
y = 2*ln(y) - 1
y - 2*ln(y) + 1 = 0 <--- stuck here

Any help would be appreciated

 

[SammyS]

Homework Statement

Hi
I am having trouble solving for y for the following equation:
e^(y) = y^(2) - 2

Homework Equations

ln(x)^(y) = y*ln(x)
ln(e^(y)) = y
ln(2) = 1

The Attempt at a Solution

My attempt:
e^(y) = y^(2) -2
ln(e^(y)) = ln(y^(2)) - ln(2)
y = 2*ln(y) - 1
y - 2*ln(y) + 1 = 0 <--- stuck here

Any help would be appreciated

Hello flyingfishcattle .

You have some mistakes:

1. ln(2) ≠ 1 . In fact, what is true: ln(e) = 1, also ln(2) ≅ 0.693 as well as log₂(2) = 1 .
2. In general, ln(A − B) ≠ ln(A) − ln(B) . Specifically, ln(y^(2) − 2) ≠ ln(y^(2)) − ln(2) .

Equations such as this cannot be solved in closed form using standard functions. This equation should be solved using something like numerical or graphical methods.

 

[YoungPhysicist]

https://www.desmos.com/calculator/yfsfflsspq
Two functions: $e^x$ and $x^2-2$
It should only have an approximated answer,rather than an definitive one( like $\sqrt2$).Geogebra and my graphing calculator also says that.

 

[LCKurtz]

A decimal approximation is -1.491644195.

 

[DuckAmuck]

This cannot be solved analytically.
You have to use an approximation.

For example, you can take the approximation (taylor series) of e^y = 1 + y + y^2/2
Then you will have 0 = y^2 - 2y - 6, and can solve for y.
You will get -1.6458 which is approaching the correct answer. To get more correct, include more terms from the taylor series.

 

[LCKurtz]

This cannot be solved analytically.
You have to use an approximation.

For example, you can take the approximation (taylor series) of e^y = 1 + y + y^2/2
Then you will have 0 = y^2 - 2y - 6, and can solve for y.
You will get -1.6458 which is approaching the correct answer. To get more correct, include more terms from the taylor series.

If you are going to do it essentially "by hand", it's much more efficient to just use a couple of iterations of Newton's method.

 

[neilparker62]

If you are going to do it essentially "by hand", it's much more efficient to just use a couple of iterations of Newton's method.

The first approximation using the 3 term Taylor expansion will be 1 - √7 which you can punch into the calculator. Thereafter you can very easily set up the Newton Rhapson iteration on any scientific calculator having an 'ANS' button (just about all of them!):

$$ANS - \frac{e^{ANS}-{ANS}^2+2}{e^{ANS}-2 \times ANS}$$

Just keep pressing the "=" sign for continuing iterations!