Question

I need help with this question

Answer 1

Below :)

Explanation:

Here u = 2^y

• To Solve:

2^(y+3) +2^-y = 6

• Given can be rewritten as:

2^y * 2³ +1/2y = 6

• Putting u = 2^y
• u*8 +1/u = 6

Actually here we multiplied u on second number with 8u and thus it results in 8u*u = 8u²[Simple LCM Trick]. Then, on LhS WE GET, (8u²+1)/u= 6

• Taking u on RHS,
• 8u² +1 = 6u
• 8u² +1 -6u = 0

Arranging terms,

• 8u² -6u +1 = 0

is in form of ax² +bx + c = 0

by mid term factor,

• 8u² -4u -2u +1 = 0
• 4u(2u -1) -1(2u-1) = 0
• Taking common,
• (2u-1) (4u-1) = 0

Either u = 1/2

• Or, u = 1/4

Hence u = 2^y has value of y as;

• 1/4 = 2^y
• (1/2)² = 2^y
• 2^-2 = 2^y[Bases are same.]

y = -2

OR,

• 1/2 = 2^y
• 2^-1 = 2^y
• y = -1[Since bases are same]