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

Question

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.

Answer 1

Case 1. One root - function strict increases

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

Answer 2

$f(x)=3^{|x|}-|2-|x||$ 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 $g(x)$