Solving logarithm problem for Y

[fr33pl4gu3]

(For those values of x for which a solution exists), solve the following equation for y

3e^3y-6 = 2x²-1

What does it mean by the values of x exists??

 

[Hootenanny]

(For those values of x for which a solution exists), solve the following equation for y

3e^3y-6 = 2x²-1

What does it mean by the values of x exists??

Because of the nature of the domain of the logarithm one can not evaluate y=\ln x for all x. Specifically the domain of the logarithm is all positive numbers. Therefore, we say that no solution exists for y=\ln x in the domain x\in\left(-\infty, 0\right].

So the question is asking you to solve the equation for y, for all values of x which exist. I hope that makes sense.

 

[Swerting]

(For those values of x for which a solution exists), solve the following equation for y

3e^3y-6 = 2x²-1

What does it mean by the values of x exists??

Say you have ln(x)=y.
This could be seen as e^y=x
It is impossible to raise any real number to any power and have it equal 0 or any number below that (feel free to try, in fact, I encourage it). Since it can't exist as x\in(-\infty,0), there are only certain numbers it can exist as.

 

[tiny-tim]

(For those values of x for which a solution exists), solve the following equation for y

3e^3y-6 = 2x²-1

What does it mean by the values of x exists??

Hi fr33pl4gu3!

Simple answer:

3e^3y-6 can only be positive.

So the equation doesn't work if 2x²-1 is negative or zero.