Number of real roots

Question

Find number of real roots of the equation

$$3^{|x|}-|2-|x||=1$$

My try:I have tried to remove the modulas by assuming x in some intervals and moved the linear part to RHS and drawn the rough graph of LHS and RHS and tried to interpret if there would be any intersection of two graphs but i could only interpret there are 1 solutions but it is not correct.

Is there some general way to solve such questions and how to solve this question without using any graphing tool ,we just have a pen and paper to solve it.

Answer 1

I think graphing would be the best option, the graph below shows the curves $3^{|x|}$ in blue and $1 + |2-|x||$ in red, they intersect in two places, so your equation has two real solutions.

It is quite easy to sketch these graphs. $3^{|x|}$ is simply an exponential function that goes to infinity fast after $x>1$ and you simply reflect the shape in the $y$-axis for $x<0$. There is a $y$-intercept of $1$.

On the other hand $1 + |2-|x||$ is simply a reflection in the $x$-axis of $|x|$, then translated upwards by $2$. Then you take the absolute value of that and translate it upwards by $1$ unit. The $y$-intercept is $3$.

Answer 2

$|2-|x||=3^{|x|}-1$

Case $\#1:$

Clearly, $x=0$ does not satisfy the equation

Case $\#2:$

If $x>0,|2-x|=3^x-1$

Case $\#2A:$ If $x-2\ge0\iff x\ge2, x-2=3^x-1\ge8\iff x\ge9$

$\implies3^x-1>x-2$

Case $\#2B:$ If $0<x<2,2-x=3^x-1\ge0\iff x\le2$

Now $2-x$ is decreasing and $3^x-1$ is increasing

At $x=0,2-0=2,3^0-1=0$

At $x=2,2-2=0,3^2-1=8$

So, there must be a root in $(0,2)$

Case $\#3:$

If $x<0, |2+x|=3^{-x}-1$

Set $-x=y, y>0$ to get $|2-y|=3^y-1$ which is same as $\#2$

Answer 3

By symmetry, if $x$ is a root, so is $-x$. Let us assume $x>0$ and solve

$$3^x-|2-x|=1,$$ i.e for $$x<2,\ 3^x-x-3=0$$ and for $$x>2,\ 3^x+x+1=0.$$ The latter relation is not possible, by positivity of the terms.

As $(3^x-x-3)'=\ln(3)3^x-1$, the function is increasing and has at most one real root. There is indeed one in the interval $(1,2)$, as the function goes from $-1$ to $4$.

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More precisely, with the Lambert function

$$x=\pm\left(3+\frac{W\left(-\frac{\ln(3)}{27}\right)}{\ln(3)}\right).$$