# Isolate y from e

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1. Oct 5, 2014

### Mrencko

1. The problem statement, all variables and given/known data
X=e^y i just need to do that i think this is maybe log but i dont know

2. Relevant equations

3. The attempt at a solution

2. Oct 5, 2014

### LCKurtz

Show us what happens if you take the natural logarithm of both sides.

3. Oct 5, 2014

### Mrencko

Log(x) =ylog(e)???

4. Oct 5, 2014

### SteamKing

Staff Emeritus
But what is the natural log of e? Think about what a natural logarithm is, and how it's defined.

5. Oct 5, 2014

### Mrencko

Its one, can you apoint me to the solution i am lost

6. Oct 5, 2014

7. Oct 5, 2014

### Mrencko

I found the solution y=-log(x)

8. Oct 5, 2014

### Mrencko

Sorry its y=-lnx

9. Oct 5, 2014

### Staff: Mentor

Neither one of these is correct. Please show what you did to get your last equation.

10. Oct 5, 2014

### Mrencko

X=e^-y then lnx=-ylne then (lnx=-y)-1.... Then (y=-lnx) or y=ln(1/x)

11. Oct 5, 2014

### Matterwave

Your original post had the equation x=e^y rather than x=e^-y. Which problem are you doing? The solution you have here is correct for x=e^-y except I'm not sure why you have a -1 in the 3rd equation which disappears in later equations yielding the correct answer.

12. Oct 5, 2014

### SteamKing

Staff Emeritus
If log (e) = 1, then what is y log (e)?

13. Oct 5, 2014

### Mrencko

Sorry i forgot the - and the-1 its just to change the - x to x and i was wrong whit log its ln

14. Oct 5, 2014

### Staff: Mentor

If the original equation was supposed to be x = e-y, then the equivalent equation is y = -ln(x). By "equivalent" I mean that every ordered pair (x, y) that satisfies the first equation also satisfies the second equation.

15. Oct 5, 2014

### Mrencko

Yes thanks dude, i really apreciate the help