Question

log2x+log4x+log16x=49/4 what is the value of x​

Answer 1

Required Answer:-

Correct Question:

• ㏒₂(x) + ㏒₄(x) + ㏒₁₆ = 49/4

To Find:

• The value of x.

Solution:

Given that,

→ ㏒₂(x) + ㏒₄(x) + ㏒₁₆ = 49/4

Using formula, ㏒ₓ(y) = ㏒(y)/㏒(x), we get,

→ ㏒(x)/㏒(2) + ㏒(x)/㏒(4) + ㏒(x)/㏒(16) = 49/4

→ ㏒(x)/㏒(2) + ㏒(x)/㏒(2²) + ㏒(x)/㏒(2⁴) = 49/4

→ ㏒(x)/㏒(2) + ㏒(x)/[2 ㏒(2)] + ㏒(x)/[4 ㏒(2)] = 49/4

LCM of ㏒(2), 2 ㏒(2) and 4 ㏒(2) is - 4 ㏒(2),

→  [4 ㏒(x) + 2 ㏒(x) + ㏒(x)][4 ㏒(2)] = 49/4

→ [7 ㏒(x)]/[4 ㏒(2)] = 49/4

→ 7/4 × ㏒(x)/㏒(2) = 49/4

→ ㏒(x)/㏒(2) = 49/4 × 4/7

→ ㏒₂(x) = 7

→ x = 2⁷

→ x = 128

→ Hence, the value of x is 128.

Answer:

• x = 128.

Logarithm Formulae:

• ㏒₁₀(x) is same as ㏒(x)
• ㏒ₑ(x) is same as ㏑(x)
• ㏒ₓ(y) = ㏒(y)/㏒(x)
• ㏒(x) + log(y) + ... = ㏒(xy...)
• ㏒(x) - ㏒(y) = ㏒(x/y)
• ㏒(xʸ) = y ㏒(x)
• ㏒(1) = 0
• ㏒ₓ(x) = 1  (x ≠ 1)
• xˡᵒᵍ ᵇᵃˢᵉ ˣ⁽ᵏ⁾ = k

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