1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.