Examine the number of solutions depending on the parameter

Question

Examine the number of solutions depending on the parameter $m$

$2^{|x-2|+x}=m^2$

I know how looks plot of $2^{|x-2|+x}$. I know how to do it when it would be $m$ instead of $m^2$

Answer 1

There is two cases:

a) x > 2 -> |x-2| = x-2:

$2^{2x-2} = m^2$

$2x-2 = \log _2 m^2$

$2x-2 = 2*\log _2 m$

$x = \log _2 m + 1$

b) x <= 2 -> |x-2| = 2-x:

$2^2 = m^2$

$m=\pm 2$

So, for $m=\pm2$ any $x<=2$ will fit, and for other nonzero m there is one solution (pay attension that m equivalent to -m, since it squared). For $m=0$, we have $2^{|x-2|+x}=0$ has no solution.

Edit:

as noted below, in part a) I forgot:

We have $x > 2$ and $x = \log _2 |m| + 1$, what means $log _2 |m| + 1 > 2$, and therefore $log _2 |m| > 1$ and $|m|>2$. For $|m|<2$ there is no solution.