Question

Find the value of log base 2 [log base 2 {log base 3 ( log base 3 27³)}]

Answer 1

Required Answer:-

Given To Evaluate:

• ㏒₂[㏒₂{㏒₃(㏒₃27³)}]

Solution:

We know that,

→ ㏒ₓyⁿ = n ㏒ₓy

So,

㏒₂[㏒₂{㏒₃(㏒₃27³)}]

= ㏒₂[㏒₂{㏒₃( ㏒₃(3³)³)}]

= ㏒₂[㏒₂{㏒₃(㏒₃3⁹)}]

= ㏒₂[㏒₂{㏒₃(9 ㏒₃3)}]

As ㏒ₓx = 1 (x ≠ 1), so we get,

= ㏒₂[㏒₂{㏒₃(9)}]

= ㏒₂[㏒₂{㏒₃(3)²}]

Again, applying the same formula, we get,

= ㏒₂[㏒₂{2 ㏒₃3}]

= ㏒₂[㏒₂{2}]

= ㏒₂[1]

As 2⁰ = 1, therefore,

→ ㏒₂(1) = 0

★ So, the final result obtained is - 0.

Answer:

• ㏒₂[㏒₂{㏒₃(㏒₃27³)}] = 0

Logarithms Formula:

• n = ㏒(x) or 10ⁿ = x.
• ㏒(x) is same as ㏑(x)
• ㏒ₓ(1) = 0, (x ≠ 0, 1)
• ㏒ₓ(a/b) = ㏒ₓ(a) - ㏒ₓ(b), a/b>0
• ㏒ₓ(ab...) = ㏒ₓ(a) + ㏒ₓ(b) +...
• ㏒ₓ(1/n) = -㏒ₓ(n)
• ㏒ₓyⁿ = n ㏒ₓy

•••♪

Answer 2

Answer:

ans: 0

Step-by-step explanation:

$\log_{2} [\ log_2 ( \log_3 ( log_3(27^3))]\\\\=\log_{2} [\ log_2 ( \log_3 ( log_3((3^3)^3))]\\\\=\log_{2} [\ log_2 ( \log_3 ( log_3(3^9)]\\\\=\log_{2} [\ log_2 ( \log_39)]\\\\=\log_{2} [\ log_2 ( \log_33^2)]\\\\=\log_{2} [\ log_22]\\\\\bf =\log_2[1]\\\\\bf=0$