Question

find the set of all solutions of the equation 2^|y| -|2^(y-1) -1|= 2^(y-1) +1........ plz explain the answer.
​

Answer 1

Here, 2

∣y∣

−

∣

∣

∣

2

y−1

−1

∣

∣

∣

=2

y−1

+1

We know to define modulus, we have three cases as

Case I: y< 0

⇒ 2

−y

+(2

y−1

−1)=2

y−1

+1

⇒ 2

−y

=2

1

(as when y< 0 |y|=-y and

∣

∣

∣

2

y−1

−1

∣

∣

∣

=−(2

y−1

−1))

Hence, y=-1, which is true when y< 0 (i)

Case II: 0≤y<1

⇒ 2

y

+(2

y−1

−1)=2

y−1

+1

⇒ 2

y

=2 (as when 0≤y<1 |y|=-y and

∣

∣

∣

2

y−1

−1

∣

∣

∣

=−(2

y−1

−1))

⇒ y=1, which shows no solution as,

0≤y<1 (ii)

Case III: y≥1

⇒ 2

y

−(2

y−1

−1)=2

y−1

+1

⇒ 2

y

=2

y−1

+2

y−1

⇒ 2

y

=2.2

y−1

(as when y≥0 |y|=y and

∣

∣

∣

2

y−1

−1

∣

∣

∣

=(2

y−1

−1))

⇒ 2

y

=2

y

, which is an identity therefor, it is true ∀y≥1 (iii)

Hence, from Eqs. (i), (ii), (iii) the solution of set is {y:y≥1∪y=−1}.