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Find the number of roots of the equation $3^{|x|}-|2-|x||=1$

My working:

Let $t$ be any positive real number.
$3^{t}-|2-t|=1$

Case 1:
$t<2$
$3^{t}-2+t=1$
$3^{t}+t=3$

Case 2:
$t>2$
$3^{t}+2-t=1$
$3^t+1=t$

Now I don't know how to proceed further to solve these equations. I would require a hint for that.

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    First suggestion would always be to plot a graph of f(t) = 0 and see the number of intersections with the x-axis – 2017-02-24
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    @mrnovice Okay...any algebraic method? – 2017-02-24
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    The more rigorous way would be to prove that $3^{t}+1$ is strictly increasing, then show that $3^{t} +1 > t \forall t$ therefore no solutions in case 2. In case 1 take a similar approach to see that there must be 1 solution. – 2017-02-24
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    This is an even function. So it suffices to consider the case when $ \ge 0$. You will find it has one root somewhere between 0 and 1. Hence there are two roots for the equation – 2017-02-24

2 Answers 2

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Case 1. One root - function strict increases

Case 2. $3^t-t$ is strictly insreases, no roots

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    Thank you! What if I had to find the roots? – 2017-02-24
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    This seems incorrect. The function is an even function. So it will have two roots (because you found one) – 2017-02-24
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    It is not possible in elementary functions – 2017-02-24
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    There is a third case when t<0 – 2017-02-24
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    @Shailesh One root in Case1 is two roots of the base fucntion – 2017-02-24
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    @mrnovice t=|x|, it is always non-negative – 2017-02-24
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    Yes but as Shailesh pointed out, there would be two roots, since you haven't accounted for when t<0 – 2017-02-24
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    @mrnovice how can t be less than zero when t=|x| – 2017-02-24
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    Oh my mistake, ignore all previous comments lol. – 2017-02-24
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$f(x)=3^{|x|}-|2-|x||$ is an even function ,so

$f(x)=1$ has symmetric roots (like $x=\pm a$)

so ,solve the equation for $x \geq 0 $ and then add negative of root(s) $$3^{|x|}-|2-|x||=1, x \geq 0 \\ 3^x-|2-x|=1 \to \begin{cases}3^x-(2-x)=1 & x <2\\3^x+(2-x)=1 & x >2\end{cases}\\ \begin{cases}3^x=3-x & x <2\\3^x=-1-x & x >2\end{cases}\\ \begin{cases}3^x=3-x & x <2\\3^x=-1-x & x >2 \end{cases}$$ first one has a solution , but second one has not root because $x>2 \to -1-x<-3\\3^x>0$

finally the original equation has 2 roots

another observation is to take $3^{|x|}=1+|2-|x||$ there is $f(x)=g(x)$ plot them together ,so must be there two cross section between $f(x)$ and $g(x)$

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