Inverse ln function

  • #1
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Homework Statement



Determine[tex] f^{-1}(0)[/tex] of [tex] f(x) = ln (4-7x) + ln(-7-5x)[/tex]

Homework Equations



[tex]f(x) = ln (4-7x) + ln(-7-5x)[/tex]


The Attempt at a Solution



What I did was
[tex]y = ln (4-7x)(-7-5x)[/tex]

Then I swapped x and y to attempt to make the inverse function

[tex]x = ln (4-7y)(-7-5y)[/tex]

and I start to solve for y

[tex]e^{x} = 35y^2 + 29y -28[/tex]

But this is where I am stuck. What do I do now? :/
 

Answers and Replies

  • #2
fzero
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Do you remember the quadratic formula?
 
  • #3
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Do you remember the quadratic formula?

Yeah. But I don't think it would work here. Would it?
 
  • #4
fzero
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Yeah. But I don't think it would work here. Would it?

You have to be careful because only one of the roots is compatible with the domain on which f(x) is defined.

Also, I just realized you swapped x and y. Don't do this as you're just going to confuse yourself. Just write

[tex] y =f(x), x = f^{-1}(y)[/tex]

and solve for x.
 
  • #5
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You have to be careful because only one of the roots is compatible with the domain on which f(x) is defined.

Also, I just realized you swapped x and y. Don't do this as you're just going to confuse yourself. Just write

[tex] y =f(x), x = f^{-1}(y)[/tex]

and solve for x.

Yeah, the second part is undefined as you cannot take the ln of a negative number. So should I discard that root and write [tex] y = ln (4-7x) [/tex] and solve for x out of here?
 
  • #6
fzero
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Yeah, the second part is undefined as you cannot take the ln of a negative number. So should I discard that root and write [tex] y = ln (4-7x) [/tex] and solve for x out of here?

By roots I meant the solutions of the quadratic equation you get, not the terms in f(x). You can't change f(x), but you can determine the range of x for which both terms in f(x) are well-defined. Remember that x doesn't have to be positive, just the argument of the log function does.
 
  • #8
fzero
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I'm confused.

Can you write down an inequality on x such that

[tex]f(x) = ln (4-7x) + ln(-7-5x)[/tex]

is well-defined?
 
  • #9
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Can you write down an inequality on x such that

[tex]f(x) = ln (4-7x) + ln(-7-5x)[/tex]

is well-defined?

x < -5/7?
 
  • #10
fzero
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x < -5/7?

It's actually x < - 7/5. Now find the roots of the quadratic equation

[tex]e^{y} = 35x^2 + 29x -28[/tex]

and determine if either of them are consistent with that inequality.
 
  • #11
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It's actually x < - 7/5. Now find the roots of the quadratic equation

[tex]e^{y} = 35x^2 + 29x -28[/tex]

and determine if either of them are consistent with that inequality.

But it's not equal to zero :/. I would just get -1.4 and 4/7.
 
  • #12
fzero
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But it's not equal to zero :/. I would just get -1.4 and 4/7.

Rewrite it in the form

[tex] 35x^2 + 29x -(e^{y}+28) =0. [/tex]
 
  • #13
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Rewrite it in the form

[tex] 35x^2 + 29x -(e^{y}+28) =0. [/tex]

What the? I never even knew that's possible.

I just sub that into the quadratic formula with [tex] c = (e^{y}+28) [/tex]
 
  • #14
fzero
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What the? I never even knew that's possible.

I just sub that into the quadratic formula with [tex] c = (e^{y}+28) [/tex]

Well we're considering x as a function of y, so x(y). As long as the coefficients depend on y and not explicitly on x, it's still a quadratic equation in x.
 
  • #15
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ugh this is so confusing :/
 
  • #16
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Is the answer -1.414?
 
  • #17
fzero
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  • #18
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I have another question of inverse ln.

Let's say you have [tex]-(ln (x)+2)^{1/2}[/tex]

The domain would just be [tex]0 < x </= e^2[/tex]

The inverse of it would be [tex] x = e^{(-y^2 + 2)}[/tex]

Would the domain of the inverse function just be x < 0? which is the range of the original function.
 
  • #19
fzero
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I have another question of inverse ln.

Let's say you have [tex]-(ln (x)+2)^{1/2}[/tex]

The domain would just be [tex]0 < x </= e^2[/tex]

The inverse of it would be [tex] x = e^{(-y^2 + 2)}[/tex]

Would the domain of the inverse function just be x < 0? which is the range of the original function.

The domain of y(x) is just [tex]x>0[/tex]. The inverse function is positive for all y (including imaginary y), so there's no problem.
 
  • #20
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The domain of y(x) is just [tex]x>0[/tex]. The inverse function is positive for all y (including imaginary y), so there's no problem.

Explain.

when x = 8, it doesn't exist.
 
  • #21
fzero
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Explain.

The log requires x>0. The square root is real whenever its argument is positive, imaginary if the argument is negative. The exponential is defined over the real axis, because of the square in its argument it gives a real even when y is imaginary. x(y) is defined everywhere that y(x) and vice-versa.
 
  • #22
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The log requires x>0. The square root is real whenever its argument is positive, imaginary if the argument is negative. The exponential is defined over the real axis, because of the square in its argument it gives a real even when y is imaginary. x(y) is defined everywhere that y(x) and vice-versa.

I'm going to presume and say I am wrong. Now, I am more confused!
 
  • #23
HallsofIvy
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Is the answer -1.414?
The answer to what question? You originally asked for the inverse function. Do you think the inverse function is "-1.414"?

You have reduced your problem to the equation
[tex] 35y^2 + 29y -(e^{x}+28) =0[/tex]
(I have swapped x and y as you did before) and you want to solve for y.

The quadratic formula says that solutions to the equation ay^2+ by+ c= 0 are given by [tex]\y= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}[/tex]

Here, a= 35, b= 29, and [itex]c= -(e^x+ 28)[/itex]. Since c involves the variable x, so will the solution to that quadratic equation. You will need to check to see whether the "+" or "-" in "[itex]\pm[/itex]" is proper for this problem.
 
  • #24
HallsofIvy
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I have another question of inverse ln.

Let's say you have [tex]-(ln (x)+2)^{1/2}[/tex]

The domain would just be [tex]0 < x </= e^2[/tex]
In order that we be able to take the logarithm, we must have x> 0. In order to be able to take the square root, we must have [itex]ln(x)+ 2\ge 0[/itex] or [itex]ln(x)\ge -2[/itex]. Since ln is an increasing function that is the same as saying [itex]x\ge e^{-2}[/itex]. That is NOT the same as [itex]x\le e^2[/itex]! Since [itex]e^{-2}> 0[/itex], we have both [itex]x> 0[/itex] and [itex]x\ge e^{-2}[/itex] just by taking [itex]x\ge e^{-2}[/itex]. That is the domain.

The inverse of it would be [tex] x = e^{(-y^2 + 2)}[/tex]
Be careful about writing x as a function of y. If y= 3x, then x= y/3 is the same function, not the inverse. The inverse function to y= 3x is y= x/3.

The original function is given by [itex]y= -(ln(x)+ 2)^{1/2}[/itex]. The inverse function is given by [itex]x= -(ln(y)+ 2)^{1/2}[/itex] but we want to write "y= " so we need to solve for y. Squaring both sides gives [itex]x^2= ln(y)+ 2[/itex] so that [itex]ln(y)= x^2- 2[/itex] and then [itex]y= e^{x^2- 2}[/itex]. The "-" in the original disappeared when we squared both sides. Stictly speaking, the natural domain of [itex]y= e^{x^2- 2}[/itex] is "all real numbers" but this is NOT that function:

Would the domain of the inverse function just be x < 0? which is the range of the original function.
Yes, exactly! When [itex]x= e^{-2}[/itex] in the original function, y= 0. As x increases without bound, ln(x) increases without bound and so does the square root of ln(x)+ 2. But y is the negative of that. The range of the original function is "all x< 0" and so the domain is also, just as you say! (A function and its inverse "swap" range and domain just as the "swap" x and y.)
 
  • #25
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So my domain for the original function is wrong too? How? It cannot go past e^2 or otherwise you'd have a negative square root.
 

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