Question

2^x=4^y=8^z and xyz=288 find 1/2x+1/4y+1/6z

Answer 1

The answer is given below :

We consider the base of the log as e.

Given that,

2^x = 4^y = 8^z = e^k (say), where k ≠ 0

Now, taking log, we get

log (2^x) = log (4^y) = log (8^z) = log (e^k),
since log(a^b) = b × loga

⇒ x log2 = y log4 = z log8 = k loge

⇒ x log2 = y log4 = z log8 = k,
since loge = 1

So, x log2 = k ⇒ x = k/log2

y log4 = k ⇒ y = k/log4

z log8 = k ⇒ z = k/log8

Also, given

xyz = 288

⇒ k/log2 × k/log4 × k/log8 = 288

⇒ k × k × k = 288 × log2 × log4 × log8

⇒ k³ = 288 × log2 × log(2²) × log(2³)

⇒ k³ = 288 × log2 × 2 × log2 × 3 × log2,
since log(a^b) = b × loga

⇒ k³ = 1728 × (log2)³

⇒ k³ = (12 × log2)³

So, k = 12 × log2

Then,

x = k/log2 = (12 × log2)/log2 = 12

y = k/log4 = (12 × log2)/(2 × log2) = 6

z = k/log8 = (12 × log2)/(3 × log2) = 4,
since log4 = 2 × log2, log8 = 3 × log2

Now, 1/(2x) + 1/(4y) + 1/(6z)

= 1/(2 × 12) + 1/(4 × 6) + 1/(6 × 4)

= 1/24 + 1/24 + 1/24

= 3/24

= 1/8

Thank you for your question.