e^-x, solve for x

[ZedCar]

1. The problem statement, all variables and given/known data
Solve for x

0.5y = e^-x

3. The attempt at a solution

I believe the answer is x = ln(2/y)

Would anyone be able to explain the intermediate steps please?

If I was trying it I would have got -ln0.5y = x

[Curious3141]

1. The problem statement, all variables and given/known data
Solve for x

0.5y = e^-x

3. The attempt at a solution

I believe the answer is x = ln(2/y)

Would anyone be able to explain the intermediate steps please?

If I was trying it I would have got -ln0.5y = x

Your answer is right, and you can easily rearrange it to ln (2/y).

What is the relationship between -ln (a) and ln (a)?

[cbetanco]

If you take the log of either side you get

ln(y/2)=-x or rearranging terms and using ln(a/b)=ln(a)-ln(b) gives x=-ln(y/2)=-(ln(y)-ln(2))=ln(2/y)

[Mentallic]

If you take the log of either side you get

ln(y/2)=-x or rearranging terms and using ln(a/b)=ln(a)-ln(b) gives x=-ln(y/2)=-(ln(y)-ln(2))=ln(2/y)

Or more simply,

a\cdot \ln(b)=\ln(b^a)

so

-\ln(y)=\ln(y^{-1})=\ln(1/y)

[ZedCar]

Thanks very much guys! :smile:

[Mark44]

Or more simply,

a\cdot \ln(b)=\ln(b^a)

so

-\ln(y)=\ln(y^{-1})=\ln(1/y)

Or, -ln(y) = 0 - ln(y) = ln(1) - ln(y) = ln(1/y).

Here, I'm using the property that ln(a/b) = ln(a) - ln(b) (in reverse).