If x is a positive real number such that 4log₁₀⁡x+ 4log₁₀₀⁡x+8log_1000⁡x=13, then the greatest integer not exceeding x, is

The correct answer is 31.

To find the value of $x$ and then its greatest integer function, we start with the given logarithmic equation:
$4{\log }_{10}x+4{\log }_{100}x+8{\log }_{1000}x=13$

First, we can express all the bases of the logarithms in terms of base 10:
100 = 10²
1000 = 10³

Using the base-change property of logarithms, specifically ${\log }_{b^k}a=\frac{1}{k}{\log }_{b}a$, we can rewrite the terms as follows:
For the second term:
$4{\log }_{100}x=4{\log }_{{10}^2}x=4·\frac{1}{2}{\log }_{10}x=2{\log }_{10}x$
For the third term:
$8{\log }_{1000}x=8{\log }_{{10}^3}x=8·\frac{1}{3}{\log }_{10}x=\frac{8}{3}{\log }_{10}x$

Now, substitute these back into the original equation:
$4{\log }_{10}x+2{\log }_{10}x+\frac{8}{3}{\log }_{10}x=13$

Combine the terms by factoring out ${\log }_{10}x$:
$(4+2+\frac{8}{3}){\log }_{10}x=13$
Simplify the sum inside the parenthesis:
$6+\frac{8}{3}=\frac{18}{3}+\frac{8}{3}=\frac{26}{3}$
So, the equation becomes:
$\frac{26}{3}{\log }_{10}x=13$

Solve for ${\log }_{10}x$ by multiplying both sides by $\frac{3}{26}$:
${\log }_{10}x=13·\frac{3}{26}$
${\log }_{10}x=\frac{3}{2}$

Converting the logarithmic form to its exponential equivalent:
$x={10}^{3/2}$
$x={\left({10}^3\right)}^{1/2}=\sqrt{1000}$

Now we need to find the greatest integer not exceeding $x$ (which is $\left[x\right]$):
Let us find the perfect squares close to 1000:
31² = 961
32² = 1024
Since $961<1000<1024$, taking square roots gives:
$31<\sqrt{1000}<32$
Therefore, the greatest integer not exceeding $x$ is:
$\left[x\right]=31$