how do you do this logarithm?

Question

y = e^x +2

the answer is y = ln x - 2 but how do you get the answer and work it out?

Answer 1

I'm not really sure if your formulas are OK. I guess you meant to say x = ln(y - 2) in the last one. I guess this is what you want:
y = e^x + 2
y - 2 = e^x
ln(y - 2) = ln e^x
ln(y - 2) = x

Answer 2

I think there's something wrong here !

If y = e raised to the power ( x + 2 ) , then , by taking the natural logarithm on both sides , we get ln(y) = x + 2.

If y = e raised to the power x , plus 2 , then by shifting 2 to the left hand side , we get y - 2 = e^x. Now taking the natural logarithm on both sides , we get ln(y - 2) = x.

Either way , we do not get y = ln(x) - 2 !

Answer 3

To get that answer, it should be to find the inverse of the equation, not solve.
If your question is:
1. y = e^(x+2)
log both sides: ln y = x+2 ; x = ln y - 2
ie, y = ln x - 2

2. y = e^x + 2 -> y - 2 = e^x
log both sides: ln (y-2) = x
ie, y = ln (x-2)

Answer 4

What exactly do you want to find. if it is y it is already given y=e^x+2. If you need make x the subject subtract 2 both sides and take log both sides you get
x=ln(y-2)
You cannot solve one equation if it has two vatiables.

Answer 5

If you are asking for the inverse function
y-2 = e^x and ln(y-2)=x
Interchange x and y and you get y=ln(x-2)

Answer 6

select one of the forms ,
y = 2*(e^x)*cosh(x)
x = ln(y-2)