If x is a positive real number such that 4log₁₀x + 4log₁₀₀x + 8log₁₀₀₀x = 13, then the greatest integer not exceeding x, is

Question

UPSC Quizroom · Accepted Answer

Step 1: Express all logarithms in base 10

log₁₀₀ x = log₁₀ x / log₁₀ 100 = log₁₀ x / 2

log₁₀₀₀ x = log₁₀ x / log₁₀ 1000 = log₁₀ x / 3

Plug in:

4·log₁₀x + 4·(log₁₀x / 2) + 8·(log₁₀x / 3) = 13

Step 2: Simplify coefficients

4·log₁₀x = 4·log₁₀x

4·(log₁₀x / 2) = 2·log₁₀x

8·(log₁₀x / 3) = (8/3)·log₁₀x

Sum: 4 + 2 + 8/3 = 6 + 8/3 = 26/3

So equation:

(26/3)·log₁₀x = 13

Step 3: Solve for log₁₀x

log₁₀x = 13 × (3/26) = 39 / 26 = 1.5

Step 4: Solve for x

x = 10^1.5 = 10^(3/2) = √(10³) = √1000 ≈ 31.6227766

Step 5: Greatest integer not exceeding x

⌊x⌋ = 31

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