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I found this question in a book.

Find the value range of $p$ if interval $(1,2)$ lies between the roots of $$2^x + 2^{2-x} + p = 0.$$

The answer given is $(-\infty,-5)$. I tried solving the problem by taking the $\log$ with base $2$ of both sides, giving $$x+ 2-x + (\log_2 p)= 0$$ or $$\log_2 p = -2$$ implying $p=1/4$. However, this is not the right answer.

How should I approach this problem?

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    $\begingroup$ A sum of logarithms is not equal to a logarithm of a sum of the same terms. And a logarithm of zero (the right-hand side) does not exist. $\endgroup$ – CiaPan May 10 '18 at 11:34
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Let $z=2^x$ then the equation becomes $$Q(z):=z^2 +pz+4=0.$$ Hence the interval $(1,2)$ lies between the roots of the original equation in $x$ if and only if the interval $(2^1,2^2)=(2,4)$ lies between the roots of the second degree polynomial $Q(z)$. Since $Q$ is convex (the coefficient of $z^2$ is positive), this happens if and only if $$Q(2)=4+2p+4<0\quad\text{and}\quad Q(4)=16+4p+4<0.$$ That is $p<\min(-4,-5)=-5$.

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Writing $$2^x+\frac{4}{2^x}+p=0$$ subst6ituting $$2^x=t$$ then we get $$t^2+pt+4=0$$ solving this for $t$ we get $$t_{1,2}=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-4}$$ Can you finish? And: $$\log(a+b)\neq \log(a)+\log(b)$$

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The function $f(x)=2^x+2^{2-x}+p$ is symmetric about $x=1$ and also we note that $x=1$ is a minimum.

Now, for $(1,2)$ interval to lie between the roots, we just need $f(2) < 0$ or that $p < -5$.

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Let $t= 2^x$, so $t \in(2,4)=:I$. Now you can write $$f(t) = -t-{4\over t}$$ Clearly since $t+{4\over t}\geq 4$ so $p\leq -4$. Since $f'(t)=-1+4t^{-2}$ we see that $f'(t)<0$ for $t\in I$ so $f$ is decreasing and thus $f_{\min} = f(4) =-5$. So we have $ p\leq-5$.

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