Equation with Absolute Value Functions in the Powers

Question

I am a Grade 12 student and I was unable to even properly approach the problem given below so I would love some pointers!

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

First, I tried to assume $2^x$ as 't' and then tried to assume $2^{|x+1|}$ as t. Then, I tried squaring it up, taking logarithms on both sides but none proved fruitful. Finally, I went with graphing it which was a bit tedious and difficult without using Desmos/graphing calculators.

I would like a method that doesn't use graphing as graphing felt a bit time-consuming for me. If possible after guiding me with a method, if you could propose a more complex question in the same topic that I could solve would be highly appreciated.

Also, I am new to this forum so apologies for any errors.

Answer 1

You seem to be having trouble visualizing the three cases, as @Khosrotash has stated the process on finding the zeroes for the absolute functions and hence solving the problem, I shall limit this post as a complementary insight on how the three cases are found.

As you should be able to see from the diagram, the cases arise from finding the overlapping regions on the real number line between the possible values of $x$ from the piecewise functions of both the absolute value functions.

Answer 2

Welcome to MSE. You can take apart for $x=-1,x=1$ because of the roots $$2^{|x+1|}\to |x+1|=0 \to x+1=0\to x=-1$$ and $$|2^x-1|=0\to 2^x-1=0 \to 2^x=1 \to x= 0$$ $$\begin{cases}x<-1 & 2^{|x+1|} - 2^x = |2^x - 1| + 1\\-1 \leq x \leq 0 & 2^{|x+1|} - 2^x = |2^x - 1| + 1\\ x>0 & 2^{|x+1|} - 2^x = |2^x - 1| + 1\end{cases}$$

$$\begin{cases}x<-1 & 2^{-x-1} - 2^x = -2^x +1 + 1& x=-2\\-1 \leq x \leq 0 & 2^{x+1} - 2^x = -2^x +1 + 1& x=0\\ x>0 & 2^{x+1} - 2^x = 2^x-1 + 1 & x\in ]0,\infty[\end{cases}$$