Inverse of a natural log function.

[justchisholm]

Hey,

This is my first post and I've been running around my residence at McGill University in Montreal attempting to figure out this problem. I've asked some engineers and they're also stumped having never encountered this type of problem. I am a first year undergrad.

f(x)=ln(x)+ln(x?2), x>2
then
f?1(y)=

Is the general question. However, I have gotten further:

y=ln(x)+ln(x-2)

Step 1: Switching the x and y variables
x=ln(y)+ln(y-2)

Step 2: Using the laws of logs to combine ln
x=ln(y(y-2))

Step 3: Taking the exponent of each side to reduce the ln to one
e^x=e^(ln(y(y-2)))=y(y-2)
e^x=y(y-2)

And this is where I am stuck, isolating y such that I can complete the process.

Could someone please send me some guidance

JC

 

[mmm4444bot]

y = f(x) = ln(x) + ln(x?2) , x > 2

then

f[sup:35w18h8i]?1[/sup:35w18h8i](y) =

I'm assuming the part in red above.

It seems to me that this exericse is not asking you to determine the expression that defines the inverse of function f.

I think it wants you to answer what the output of function f[sup:35w18h8i]-1[/sup:35w18h8i](x) is when you input y = f(x).

After all, f[sup:35w18h8i]-1[/sup:35w18h8i](y) is function notation for the output of f[sup:35w18h8i]-1[/sup:35w18h8i](x) when x = y, and y is the output of function f.

If you understand the concept of an inverse function, and what it does, then the answer is obvious.

You could also include a domain statement for the function f[sup:35w18h8i]-1[/sup:35w18h8i](x), unless your teacher is a machine.

? < y < ?

 

[Deleted member 4993]

Hey,

This is my first post and I've been running around my residence at McGill University in Montreal attempting to figure out this problem. I've asked some engineers and they're also stumped having never encountered this type of problem. I am a first year undergrad.

f(x)=ln(x)+ln(x?2), x>2
then
f?1(y)=

Is the general question. However, I have gotten further:

y=ln(x)+ln(x-2)

Step 1: Switching the x and y variables
x=ln(y)+ln(y-2)

Step 2: Using the laws of logs to combine ln
x=ln(y(y-2))

Step 3: Taking the exponent of each side to reduce the ln to one
e^x=e^(ln(y(y-2)))=y(y-2)
e^x=y(y-2)

And this is where I am stuck, isolating y such that I can complete the process.

Could someone please send me some guidance

JC

Taking your problem statement literally:

e^x + 1 =y(y-2) + 1= (y - 1)[sup:3guwe2vz]2[/sup:3guwe2vz]

y = f[sup:3guwe2vz]-1[/sup:3guwe2vz](x) = - 1 ± ?(e[sup:3guwe2vz]x[/sup:3guwe2vz]+1)

Now as Mark suggested above, the domain of the original function needs to be defined - so that you choose the range of the inverse function and you would have a "legitimate function" (not one with ± sign). Anyway, then

f[sup:3guwe2vz]-1[/sup:3guwe2vz](y) = - 1 ± ?(e[sup:3guwe2vz]y[/sup:3guwe2vz]+1)

.