Question

Let $x$ be a positive real number such that $4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$ , then the greatest integer not exceeding $x$. is

Answer 1

Correct Answer: 31

We are given the equation:

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

We can simplify the logarithms:

$\log_{100} x = \frac{\log_{10} x}{\log_{10} 100} = \frac{\log_{10} x}{2}$,
$\log_{1000} x = \frac{\log_{10} x}{\log_{10} 1000} = \frac{\log_{10} x}{3}$.

Substitute these into the equation:

$4\log_{10} x + 4\left(\frac{\log_{10} x}{2}\right) + 8\left(\frac{\log_{10} x}{3}\right) = 13$,
$4\log_{10} x + 2\log_{10} x + \frac{8}{3}\log_{10} x = 13$.

Factor out $\log_{10} x$:

$\left(4 + 2 + \frac{8}{3}\right)\log_{10} x = 13$,
$\left(\frac{18}{3} + \frac{8}{3}\right)\log_{10} x = 13$,
$\frac{26}{3}\log_{10} x = 13$.

Solve for $\log_{10} x$:

$\log_{10} x = \frac{13 \times 3}{26} = \frac{39}{26} = 1.5$.

Thus, $x = 10^{1.5} = 10 \times \sqrt{10} \approx 31.62$.
The greatest integer not exceeding $x$ is 31.