I need help with an exercise

[Jerson]

If x = 1-ae^2y/-be^2y Prove that y = -Ln√a-bx

 

[MarkFL]

Hello, and welcome to FMH!

First, try solving the equation for \(e^{2y}\)...what do you get?

 

[Deleted member 4993]

If I were to solve this problem, I would first substitute:

u = e^2y . and

solve for u

Which is of course essentially same as response#2.

 

[Steven G]

If x = 1-ae^2y/-be^2y Prove that y = -Ln√a-bxView attachment 17082

Please note that what you type out means the following
\(\displaystyle 1 - ae^2(\dfrac{y}{-b})e^2y\)

You need to use parenthesis!

(1-ae^(2x))/(-be^(2x))

 

[Deleted member 4993]

Please note that what you type out means the following
\(\displaystyle 1 - ae^2(\dfrac{y}{-b})e^2y\)

You need to use parenthesis!

(1-ae^(2x))/(-be^(2x)).................... why is that? The independent variable for this equation is supposed to be y

To the corner ..... on the double......

 

[firemath]

To the corner ..... on the double......

No...on the power of 5....

 

[MarkFL]

To follow up:

We are given:

[MATH]x=\frac{1-ae^{2y}}{-be^{2y}}=\frac{ae^{2y}-1}{be^{2y}}[/MATH]
This implies:

[MATH]e^{2y}=\frac{1}{a-bx}[/MATH]
Which implies:

[MATH]y=-\frac{1}{2}\ln(a-bx)=-\ln\left(\sqrt{a-bx}\right)[/MATH]