Solving equation in one unknown with absolute value

Question

Consider the following equation : $$\tfrac12e^{-|x|}=e^{-2|x-2|}$$ Now what I have done is: $$\ln\tfrac12 -|x| = -2 |x-2|$$ But I don't know how to go further because of the absolute.

Answer 1

We have $\ln\tfrac12 -|x| = -2 |x-2|$. Rearranging, we get $$\ln\tfrac12 = |x|-2 |x-2|$$.

Now we split this equation into cases and start solving.

Case $1$: $x \ge 2$, which means that all expressions in absolute value are positive, so we can remove the absolute value signs to get $$\ln\tfrac12 = x-2 (x-2)$$, which can be easily solved to show that $$x = 4 - ln(\frac12)$$. However, we specified that $x \ge 2$in this case, so we need to make sure this condition is satisfied. Using a calculator, we see that $x \approx 4.69$, so it fits and is thus a solution.

Case $2$: $ 0 \le x \le 2$, which means $x$ is positive, so we may remove the signs from $|x|$, but $x-2$ is negative, so we have to remove the signs from $|x-2|$ and negate it to make it $2-x$. Thus, we have $$\ln\tfrac12 = x-2(2-x)$$. We can easily solve thus to find $$x = {ln(\frac12)+4 \over 3}$$. A quick calculation tells us that $x \approx 1.1$, which is in the domain we specified in this case, and is thus a solution.

Case $3$: $x \le 0$, all absolute values are negative, so we remove all the absolute value signs and negate the expressions inside them. Thus our equation becomes $$\ln\tfrac12 = -x-2(2-x)$$, which we can easily solve to find $x= 4 + ln(\frac12)$. A quick calculation tells us that $x \approx 3.3$, which unfortunately does not satisfy this case requirement of $x \le 0$, so discard this solution.

Thus, in conclusion, our solutions are $$x = 4 - ln(\frac12)$$ & $$x={ln(\frac12)+4 \over 3}$$

Answer 2

$$\ln\frac{1}{2}-|x|=-2|x-2|$$

$$\ln\frac{1}{2}=|x|-2|x-2|$$

Case 1: $x>2, |x|=x,|x-2|=x-2$

$$\ln\frac{1}{2}=-x+4$$

$$x=4+\ln(2)>2$$ so this is a solution

Case 2: $0<x<2, |x|=x,|x-2|=2-x$

$$\ln\frac{1}{2}=3x-4$$

$$0<x=\frac{\ln\frac{1}{2}+4}{3}<2$$ so this is a solution

Case 3: $0>x, |x|=-x,|x-2|=2-x$

$$\ln\frac{1}{2}=x-4$$

$$x=4+\ln\frac{1}{2}>0$$ So there is no solution given this case

Answer 3

Hint

The slow, pedantic, pedestrian method:

In a problem like this, before you attempt any math manipulations, immediately create cases to represent each relevant range of $x$.

Here, it is immediate that you have $x < 0$ and $x \geq 0.$

You also have $(x-2) < 0$ and $(x-2) \geq 0.$

So simply compress the 4 possibilities above into 3 distinct cases. Then, construe each case as a totally separate problem.

Naturally, with experience, you will look for elegant analysis/shortcuts to the slow pedantic pedestrian methods. However, when starting out, don't be seduced into trying for elegance.