If $x$ is a positive real number such that $4 \log _{10} x+4 \log _{100} x+8 \log _{1000} x=13$, t hen greatest integer not exceeding x , is

When we have logarithms with different bases, we can't directly add them. The easiest approach is to convert everything to base 10 using the change of base formula: $\log_a b = \frac{\log_{10} b}{\log_{10} a}$

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For $\log_{100} x$:

$\log_{100} x = \frac{\log_{10} x}{\log_{10} 100} = \frac{\log_{10} x}{2}$

Since $\log_{10} 100 = \log_{10} 10^2 = 2$

For $\log_{1000} x$:

$\log_{1000} x = \frac{\log_{10} x}{\log_{10} 1000} = \frac{\log_{10} x}{3}$

Since $\log_{10} 1000 = \log_{10} 10^3 = 3$

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$4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$

$4 \log_{10} x + 4 \cdot \frac{\log_{10} x}{2} + 8 \cdot \frac{\log_{10} x}{3} = 13$

$4 \log_{10} x + 2 \log_{10} x + \frac{8}{3} \log_{10} x = 13$

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Let $\log_{10} x = k$.

$4k + 2k + \frac{8}{3}k = 13$

To avoid working with fractions, we multiply everything by 3:

$3(4k) + 3(2k) + 3 \cdot \frac{8}{3}k = 3(13)$

$12k + 6k + 8k = 39$

$26k = 39$

$k = \frac{39}{26} = \frac{3}{2}$

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Since $\log_{10} x = \frac{3}{2}$, we can find $x$:

$x = 10^{3/2}$

$10^{3/2} = 10^1 \cdot 10^{1/2} = 10\sqrt{10}$

Since $\sqrt{10} \approx 3.162$:

$x = 10 \times 3.162 = 31.62$

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Since $x = 31.62$, the greatest integer not exceeding $31.62$ is $31$.

Therefore, the answer is 31.