Ratio, Proportion, Indices, LogarithmsPYQ Sept 25Question 4400 of 305
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The value of x\displaystyle x in log⁡x(4)+log⁡x(16)+log⁡x(64)=12\displaystyle \log_x(4) + \log_x(16) + \log_x(64) = 12 is

Options

A1
B2
C3
D4
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Correct Answer

✅ Option b — 2

All Options:

  • A1
  • B2
  • C3
  • D4

Detailed Solution & Explanation

Given equation: log⁡x(4)+log⁡x(16)+log⁡x(64)=12\log_x(4) + \log_x(16) + \log_x(64) = 12
We know that: 4=22,16=24,64=264 = 2^2, \quad 16 = 2^4, \quad 64 = 2^6
Using the property of logarithms log⁡b(ak)=klog⁡b(a)\displaystyle \log_b(a^k) = k \log_b(a), we can rewrite the terms as: log⁡x(22)+log⁡x(24)+log⁡x(26)=12\log_x(2^2) + \log_x(2^4) + \log_x(2^6) = 12 2log⁡x(2)+4log⁡x(2)+6log⁡x(2)=122\log_x(2) + 4\log_x(2) + 6\log_x(2) = 12
Combining the like terms on the left-hand side: (2+4+6)log⁡x(2)=12(2 + 4 + 6) \log_x(2) = 12 12log⁡x(2)=1212 \log_x(2) = 12
Dividing both sides by 12: log⁡x(2)=1\log_x(2) = 1
By the definition of logarithms, log⁡b(a)=c\displaystyle \log_b(a) = c implies bc=a\displaystyle b^c = a. Therefore: x1=2  ⟹  x=2x^1 = 2 \implies x = 2
Hence, **Option B** is the correct answer.

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