Understanding Exponentials and Logarithms: Solving Equations with ln and abs

[CaptainDunzo]

Homework Statement

I was tasked to answer/show why e^(ln(abs(y-1))=e^(x+c) simplifies to y-1=Ce^x.

Relevant Equations

none

I started with the top equation. I first said since they were both raised to e it would then change to ln(abs(y-1))=x+c. I then thought because of the abs It could be broke into 2 equations. ln(y-1)=x+c and ln(y-1)=-(x+c). I then got confused because I had 2 equations to work with instead of one.

 

[nrqed]

Homework Statement:: I was tasked to answer/show why e^(ln(abs(y-1))=e^(x+c) simplifies to y-1=Ce^x.
Relevant Equations:: none

View attachment 259134
I started with the top equation. I first said since they were both raised to e it would then change to ln(abs(y-1))=x+c. I then thought because of the abs It could be broke into 2 equations. ln(y-1)=x+c and ln(y-1)=-(x+c). I then got confused because I had 2 equations to work with instead of one.

Two points: first, the constant C in the first equation is not the same as the constant C in the second equation, so it would be better to not use the same symbol, of better write the second C in terms of the first C. You probably realize this but I thought I would mention it.

EDIT!

Oops, I wrote too quickly. You do have to consider the two cases, indeed. What will happen is that you will have two solutions. They are just being very sketchy in the way they write the equation. What they mean is that one can write $y-1 = K e^x$ for *some* constant K. The constant K will be either $e^C$ or $-e^C$, depending on the sign of $y-1$.

 

[Mark44]

Homework Statement:: I was tasked to answer/show why e^(ln(abs(y-1))=e^(x+c) simplifies to y-1=Ce^x.
Relevant Equations:: none

I started with the top equation. I first said since they were both raised to e it would then change to ln(abs(y-1))=x+c.

A better strategy would be to use the fact that $e^{\ln y} = y$ is an identity for all y > 0. You can use this fact to rearrange the left side. Another identity is $e^{a + b} = e^a \cdot e^b$.