Question

|2^x -1| + |2^x + 1| = 2 solve for x​

Answer 1

X=1/2

Explanation:

2x-1+2x+1=2

2x+2x=2

4x=2

X=2/4

X=1/2

Answer 2

Answer

x=0 and x<0

Explanation:

when x is +ve,

modulus funtion opens as it is, i.e

---> |2^x -1| + |2^x +1| = 2

---> 2^x-1 + 2^x +1 = 2

---> 2^(x+1) = 2^1

---> x+1 = 1.....Therefore x=0 is one of the solutions

Now, when x is -ve,

modulus function opens as |x| = -x

---> |2^x -1| + |2^x +1| = 2

In this part only the first term needs to be modified, because it turns to

1 - 2^x + 2^x +1 = 2....

this happens because if x is -ve in the first terms then the terms inside mod func is always less than zero so we took out the mod function and wrote it as |2^x -1| = 1-1/2^x, to make this always +ve as that is the output of a modulus function.

But in the case of the second term no matter how -ve the value of x is it will always be positive because it will be between 0 and 1 and is added to 1, so directly open it as it is, i.e solve |2^x +1| = 1/2^x +1

therefore we arrive to the following,

---> 1 - 2^x + 2^x +1 = 2

--->1+1=2

---> 2=2

LHS=RHS

Now here the solution is any negative number.

In the first part the solution was x=0, and the second part solution was x<0

Combining both of them we get

x$\leq$0

If you didn't understand the 2nd part, try substituting -ve values like, -1,-2,-3....( I understand the 2nd part was quite tough.

Please mark this as the brainliest answer.