Solving 4^x + 4^-x = 8 - 2x - x^2

[kurvmax]

Homework Statement

I have a curiosity question. The precalc textbook asks me to solve by using a graphing calculator but I want to do it algebraically. I know that the equations intersect through the graphing calculator -- one intersection at x = 1.07808, another around -1.5. The two equations are:

$$4^x + 4^{-x} = 8 - 2x - x^2$$

Homework Equations

The Attempt at a Solution

First I took the log of both sides:

$$x\ln{4} + {-x}\ln{4} = \ln{8} - \ln(2x) - \ln(x^2)$$
(not sure how to create a new line with tex, since \\ is not working for me)
$$x(\ln{4} - \ln{4}) = \ln{8} - \ln{2} + \ln{x} - \ln{x} - \ln{x}$$

Forgive my mathematical illiteracy. I sort of guessed from that I could cancel the left side since ln(4) - ln(4) = 0, but I'm probably wrong there. And anyway, what if the equation didn't work out that conveniently? If the left side can be canceled and the ln(x)es on the right-side cancel, then it seems like it could become:

$$\ln{x} = \ln{8} - \ln{2}$$

which is the same as:

$$e^{\ln{8} - \ln{2}} = x = 4$$ ??

Forgive my terrible math skills.

 

[Dick]

Relax. Not every equation can be solved algebraically. And your equation is certainly one of them. Your failure does not indicate lack of math skills, it indicates that no one could do it. But you are doing some bad stuff. E.g. log(4^x+4^(-x)) is NOT equal to x*ln(4)-x*ln(4). No way. Go with the graphing calculator solution.

 

[kurvmax]

When you take the log of something, can't you pull down the exponent as a multiplicative factor?

For example, if I was to take the log of 4^x, that would be x*log(4)?

 

[Dick]

You can. But log(a+b) is not equal log(a)+log(b). That's the mistake.

 

[kurvmax]

I see, thanks. I got log(a*b) confused with log(a+b). Which is pretty major. Actually, for some reason I thought I could take the log of each component of the two expressions individually. Is there any way to break apart things like:

$$\ln{(4^x + 4^{-x})}$$?

And when I do that log, I'm guessing that I can't pull down those xes inside. But could I change it to:
$$\ln{[(4 + 4^{-1})^x]$$

and then pull the x down?

 

[Gib Z]

Well, do you think you can pull out the exponent like that? Just try it for n=2. there's not really any "useful" way to simplifying the log of a sum.

 

[Mentallic]

I know the feeling. You're desperate enough for an answer that you begin to break mathematical rules subconsciously (even though you probably know the correct rule).

$$a^{c}+b^{c} \neq (a+b)^c$$

Usually for younger students in high school first learning to work with binomials, they make the common error of assuming the right-hand side is equal to the left.

e.g. $$(a+b)^2=a^2+2ab+b^2\neq a^2+b^2$$

 

[kurvmax]

I don't know why I thought that the exponent distributed like that ... thank you so much for your patience and help. Out of curiosity -- and at the risk of testing your patience further -- is there an easy to tell if an equation can be solved algebraically? I'm guessing the answer is when the x looks really hard to isolate?

How would I mark this thread as SOLVED?