Solve for y: X=e^yEquation Solution

[Mrencko]

Homework Statement

X=e^y i just need to do that i think this is maybe log but i don't know[/B]

Homework Equations

The Attempt at a Solution

 

[LCKurtz]

Homework Statement

X=e^y i just need to do that i think this is maybe log but i don't know[/B]

Homework Equations

The Attempt at a Solution

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

 

[Mrencko]

Log(x) =ylog(e)?

 

[SteamKing]

Log(x) =ylog(e)?

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

 

[Mrencko]

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

 

[Mark44]

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

No, we won't do your work for you, due to forum rules - https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/, in Homework Guidelines.

Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

 

[Mrencko]

I found the solution y=-log(x)

 

[Mrencko]

I found the solution y=-log(x)

Sorry its y=-lnx

 

[Mark44]

I found the solution y=-log(x)

Sorry its y=-lnx

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

 

[Mrencko]

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

 

[Matterwave]

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

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.

 

[SteamKing]

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

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

 

[Mrencko]

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

 

[Mark44]

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.

 

[Mrencko]

Yes thanks dude, i really apreciate the help