Rearrange this equation to make x the subject

autodidude

1. The problem statement, all variables and given/known data
I'm pretty sure we're meant to solve this using a calculator but I'd like to see if it can be rearranged to make x the subject. I've tried the usual approaches but ended up going in circles. It's part of a bigger problem which required me to find the intersection of [tex]f(x)=2e^x-4[/tex] and its inverse


2. Relevant equations

[tex]ln((x+4)/2) =2e^x-4[/tex]

algebrat

Haha, you're almost there, it looks like you took the right steps, you just mixed up y and x. So in a sense, you have the solving steps down, but now comes the hard part, meaning it's hard to explain.

There's two ways to go, take the original y=f(x), and solve for x in terms of y, so x=x(y). Then you trade y and x.
In your case that would look like

y=2e^x-4 solve
x=ln((y+4)/2) switch
y=ln((x+4)/2)

I think you used this method, but then set one y equal to another, which is wrong on a deep and subtle level, which if you're having trouble understanding, let us know that that is where you are stuck,a dn we should be able to go into better detail.

The other way to go, slightly different, switch y and x first then solve.
That would look like

y=2e^x-4 switch
x=2e^y-4 solve
y=ln((x+4)/2)

epenguin

If you have a graphic calculator and plot the function and its inverse: that might remind you of the diagramatic relation of a function and its inverse quite generally which you have probably been told sometime. Alternatively, work it out. Or check back in your book (probably the hardest way!).

Alternatively consider it symbolically (forget the specifics of exp and log).

You are asked x when f(x) = f^-1(x) .

What do you have to do to that equation to get x = ... ?

Mark44

It's much simpler to explain if you omit the part where you switch the names of the variables.
Since the stated goal was to solve for x in terms of y, switching variable names results in an equation where y is still in terms of x.

From the above,
y = 2e^x - 4
<=> y + 4 = 2e^x
<=> (y + 4)/2 = e^x
<=> ln[(y + 4)/2] = x

Or x = ln[(y + 4)/2]
The equation we started with is solved for x in terms of y. Note that all steps above are reversible, which is why I'm showing the equivalence symbol at each step rather than the implication symbol.

In terms of function notation, what we have here are y = f(x) and x = f^-1(y). Both of these equations have exactly the same graph. The only difference is that one equation gives you the y value if you know the x value at a point; the other gives you the x value if the y value is known.

IMO, many textbooks (and hence students who read these textbooks) are fixated on the variable name switching process, and miss out on what I believe is the more important process of solving the equation for the other variable.

The only advantage of switching variable names in the context of inverse functions is that both the original function and its inverse now have x as the independent variable and y as the dependent variable. That allows you to graph both functions on the same axis system, and observing that the graphs are the reflection of each other across the line y = x.

Personally I don't think this is a very big deal. The applications of finding inverses that come later in calculus almost never involve switching variable names.
Again, the goal was to solve for x ("make x the subject").

SammyS

Looking at this graphically:

If the function, f(x), has an inverse, and in this case it does, then the graph of y = f^-1(x) is the same as then graph x = f(y).

The graph, x = f(y), can be obtained by reflecting the graph of y = f(x) through the line, y = x.

Therefore, the graphs, y = f^-1(x), and y = f(x), intersect only where they intersect the graph of y = x

So you only need to solve, [itex]\displaystyle x=2e^{x}-4\,,\ \text{ or }x=\ln((x+4)/2)\ ,[/itex]

rather than solving [itex]\displaystyle \ln((x+4)/2)=2e^{x}-4\ .[/itex]

autodidude

Yeah, that's what I tried to do

@penguin: The answer the book gives suggests that this is a tech-active question (not a nice solution, has 3 decimal places)

You are asked x when f(x) = f-1(x) .

What do you have to do to that equation to get x = ... ?

Isn't that just rearrange for x?


At the moment, I'm most interested in seeing whether the equation can be rearranged to find x, forgetting all inverse stuff. So just pretend I'm asking how to rearrange for x :p

pcm

not necessarily, see for example,
f(x)=1/x ;x>0
f^-1(x)=1/x

and they are coincident.

dimension10

Add 4, divide by 2 and take the natural logarithm. What do you get?

SammyS

Yes, pcm, you are correct !

I suppose I must have only considered invertible functions which are increasing, when I wrote that.

There is a whole slew of invertible functions that are decreasing which are counter-examples.

autodidude

Can [tex]ln (\frac{(x+4)}{2})=2e^x-4 [/tex] even be rearranged for x? I seem to recall coming across a exponential equation in the past that I never got anywhere with. What are these equations called?

HallsofIvy

Okay, doing that you get x= ln((x+12)/2)- ln(2). How does that help?

As others have said there is NO way to solve this equation using elementary algebraic methods. In fact, autodidude himself said "The answer the book gives suggests that this is a tech-active question (not a nice solution, has 3 decimal places)".

epenguin

You didn't quite state what the question is.

You asked 'rearrange the equation to make x the subject'.

When I said do it in general independent of the nature of f I meant you can to

f^-1(x) = f(x)

apply f giving

f[f^-1(x)] = f(f(x))

so

x = f[f(x)] . RHS is a.k.a. f²(x)

You can work out f²(x) as a formula for this specific case easily.

But if you want x = something independent of x, I don't think that can be meaningfully done.

If you want to numerically find the x that satisfies the first equation above then the simplest thing is to use the symmetry as pointed out by SammyS (I was trying to lead you there too ) and solve f(x) = x .

There are two solutions (which you should expect if you sketch curves) and one is close to but slightly above -4 (which you can perhaps rationalise). On a graph plotter or something plot f(x), f^-1(x) and the line through origin of slope 1. They all go through 2 points and so does f²(x) - and lots of other related functions you can invent, if you think about it - even looks quite pretty.

autodidude

@HallsOfIvy: Ah, I must've missed it, thanks. What do you call equations that can't be solved algebraically?


Thanks epenguin and everyone else that contributed, now I can try solving it :)

jtbell

Transcendental, I think.