The question I am struggling with is asking how many unique solutions there are to the following:

$$(x-2)^2 (x+2)^2 = 4+\log(x+4)$$

Wolfram Alpha tells me the answer is four, but as far as I can figure out, you could not solve for these four solutions without a graphical calculator or from what I've gathered from Google, using the Lambert W function. But, I do not need to solve it - I just need to prove/state how many unique solutions there are.

Expanding/simplifying etc gets me to:

$$12 e^x+x = e^{12}-4$$

By looking at this, is there a way to prove this has four unique solutions? This is an A-level equivalent question so I assume the answer is not too complex, however I am stumped!

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    $\begingroup$ Show there is at least $4$. Assume there is $5$, and show that this leads to a contradiction. $\endgroup$ – Ahmed S. Attaalla Oct 9 '16 at 18:35

Hint: We rewrite the problem as finding the zeros of $(x-2)^2(x+2)^2-(4+log(x+4))$. Looking at the graph below, specifically the derivative at different regions, could you conclude that there won't be a fifth zero?

enter image description here

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    $\begingroup$ Ahh, perfect. I was thinking of it algebraically and completely overlooked the idea of just plotting both sides and seeing where they intersect! Thanks. $\endgroup$ – yerbamate170 Oct 9 '16 at 19:56
  • $\begingroup$ @yerbamate170 Is to be made more rigorous with the powers of calculus. $\endgroup$ – Simply Beautiful Art Oct 15 '16 at 12:36

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