Solve for x: 0.5y = e^-x | Intermediate Steps Explained

[ZedCar]

Homework Statement

Solve for x

0.5y = e^-x

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]

Homework Statement

Solve for x

0.5y = e^-x

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!

 

[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).