Solve for y: X=e^yEquation Solution

  • Thread starter Mrencko
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In summary, the conversation is about solving a problem involving the equation X=e^y, with the person seeking help thinking it may involve logarithms. The expert prompts them to think about what a natural logarithm is and what the solution might be. The person then realizes their mistake and finds the correct solution of y=-lnx.
  • #1
Mrencko
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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

 
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  • #2
Mrencko said:

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.
 
  • #3
Log(x) =ylog(e)?
 
  • #4
Mrencko said:
Log(x) =ylog(e)?

But what is the natural log of e? Think about what a natural logarithm is, and how it's defined.
 
  • #5
Its one, can you apoint me to the solution i am lost
 
  • #7
I found the solution y=-log(x)
 
  • #8
Mrencko said:
I found the solution y=-log(x)
Sorry its y=-lnx
 
  • #9
Mrencko said:
I found the solution y=-log(x)
Mrencko said:
Sorry its y=-lnx
Neither one of these is correct. Please show what you did to get your last equation.
 
  • #10
X=e^-y then lnx=-ylne then (lnx=-y)-1... Then (y=-lnx) or y=ln(1/x)
 
  • #11
Mrencko said:
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.
 
  • #12
Mrencko said:
Its one, can you apoint me to the solution i am lost

If log (e) = 1, then what is y log (e)?
 
  • #13
Sorry i forgot the - and the-1 its just to change the - x to x and i was wrong whit log its ln
 
  • #14
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
Yes thanks dude, i really apreciate the help
 

FAQ: Solve for y: X=e^yEquation Solution

1. What does "solve for y" mean in this equation?

"Solve for y" means to find the value of y that makes the equation true. In other words, we are looking for the value of y that satisfies the equation X=e^y.

2. How do I solve for y in this equation?

To solve for y in this equation, you can use the natural logarithm (ln) function. First, take the ln of both sides of the equation to get ln(X)=ln(e^y). Then, use the property of logarithms to rewrite the right side as y*ln(e). Finally, divide both sides by ln(e) to isolate y and get the answer y=ln(X).

3. Is there more than one solution for y in this equation?

Yes, there are infinitely many solutions for y in this equation. This is because the natural logarithm function is a one-to-one function, meaning that for every input there is only one output. However, the input X can have many different values, resulting in different outputs for y.

4. Can I solve for y if X is a negative number?

No, you cannot solve for y if X is a negative number. This is because the natural logarithm function is only defined for positive numbers. If X is a negative number, the equation X=e^y has no real solutions for y.

5. How can I use this equation in real-life applications?

This equation can be used in many real-life applications, such as calculating the growth rate of a population or the decay rate of a radioactive substance. It can also be used in finance to calculate compound interest or in physics to model exponential growth or decay. Essentially, this equation can be used whenever there is a relationship between a quantity and its rate of change over time.

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