A logarythmic/exponential equation where x is the exponent

[Ryker]

Homework Statement

2^x-2 - 2 = 5 - x

Find x.

Homework Equations

log(a)x = y

The Attempt at a Solution

I've thought about this a lot and the best I could come up with was getting to ...

x = log(2)(12 - 4x)

The problem is that I seem to be going in circles and I can't drop x to the same level on both sides of the equation. And the thing that really bothers me is that this problem isn't termed as advanced in the book that I'm going through. The book does, however, state that one should take aid in the graphs (charting of which is a part of the problem given). If you do it that way, it's pretty easy to arrive at the solution, but I want to find it without that or just randomly guessing numbers 1, 2, 3 etc. and trying them out.

Any help would be immensely appreciated :)

 

[rock.freak667]

I do not believe there is an algebraic way to solve for x. Graphing would be the simplest solution.

 

[Ryker]

There isn't? Well, if that's true, then it would make my day, because I've been trying to find a solution for hours and felt really stupid that I couldn't do a problem not even designated as "advanced" :D

But if anyone does think of a solution to this without reading the graph, I'd be much obliged.

 

[HallsofIvy]

There is no way to solve such an equation with "elementary" function though you might be able to rewrite so that you could use "Lambert's W function" which is defined to be the inverse function to f(x)= xe^x.

 

[Ryker]

Cool, that's great to hear then. Thanks again!

 

[jackmell]

I think solving this in terms of the W function offers some nice algebra practice. We wish to get it into the Lambert W form:

K=h(x)e^{h(x)}

then taking the W function of both sides we write it as:

h(x)=W[K]

where x then can be isolated from h(x). How about I just show you some of the steps and you fill in the blanks?

Begin with:

2^{x-2}=7-x

\vdots

1=-(x-7)e^{-(x-2)\log(2)}

\vdots

2^5=-(x-7)e^{-(x-7)\log(2)}

2^5\log(2)=-(x-7)\log(2)e^{-(x-7)\log(2)}

That's now in Lambert W form. Can you solve for x now?

 

[Ryker]

Hmm, are those "(2)" bases of the logarithm? If yes, which of the integers are "numbers x" of the respective logarithms? 2⁵ and -(x - 7)? Because right now, I'm still lost insofar as solving this is concerned :/

 

[jackmell]

Hmm, are those "(2)" bases of the logarithm? If yes, which of the integers are "numbers x" of the respective logarithms? 2⁵ and -(x - 7)? Because right now, I'm still lost insofar as solving this is concerned :/

In general, the symbol, "log" refers to log base e in higher math although in high-school it is usually associated with something else. So \log(2) is just log base e of 2. So starting with:

2^{x-2}=7-x

then that's just:

e^{(x-2)\log(2)}=7-x

dividing by the left-side:

1=(7-x)e^{-(x-2)\log(2)

or:

1=-(x-7)e^{-(x-2)\log(2)}

but we wish to get it into Lambert W-form so we need to make the exponent in terms of that -(x-7) term so to do that we multiply both sides by e^{5\log(2)}=2^5 to obtain:

2^5=-(x-7)e^{-(x-7)\log(2}

and then just another factor of \log(2) on both sides puts it into the desired form.

 

[HallsofIvy]

Nicely done! I (very briefly) considered trying that myself- be then decided "that's not my problem"! Glad you did it.

 

[Ryker]

In general, the symbol, "log" refers to log base e in higher math although in high-school it is usually associated with something else. So \log(2) is just log base e of 2. So starting with:

2^{x-2}=7-x

then that's just:

e^{(x-2)\log(2)}=7-x

dividing by the left-side:

1=(7-x)e^{-(x-2)\log(2)

or:

1=-(x-7)e^{-(x-2)\log(2)}

but we wish to get it into Lambert W-form so we need to make the exponent in terms of that -(x-7) term so to do that we multiply both sides by e^{5\log(2)}=2^5 to obtain:

2^5=-(x-7)e^{-(x-7)\log(2}

and then just another factor of \log(2) on both sides puts it into the desired form.

Damn, that's impressive. Though, the doofus I am, I just realized that my original equation was wrong, it was supposed to be + 2, and not - 2. So I guess I can't really say if I can solve for x after you've served it on a platter, but I was thinking one now just has to bring down e's exponent to 2^-(x-7), multiply that with -(x-7) and then carry it over to the log(2)'s exponent. After that you then solve the logarithm to get 2 on the power described above, which you in the last step equate with 5. Would that be correct?

Oh, and sorry for the late reply, the thing is I'm revising high school maths, and due to time constraints I have to move on really fast. And basically when I found out you need advanced maths to solve this problem I was relieved that at this stage I wasn't supposed to know that yet. When does one learn about Lambert's W function anyway?

 

[jackmell]

Yes, you can solve it the same way and right, you need to bring the exponent to -(x-7) or multiply by 2^9 this time. The W function is not normally studied in any course that I know of although would be most appropriately done so in Complex Analysis since it's an infinitely-valued complex-analytic function.

 

[Ryker]

Yes, you can solve it the same way and right, you need to bring the exponent to -(x-7) or multiply by 2^9 this time. The W function is not normally studied in any course that I know of although would be most appropriately done so in Complex Analysis since it's an infinitely-valued complex-analytic function.

Ah, okay, so it isn't even studied at the university level? In any case, thanks heaps for your help.