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

Given the equation: $4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$. We need to find the greatest integer not exceeding \( x \).

First, recognize the logarithmic bases in the problem. We have:

- \(\log_{100} x = \log_{10^2} x = \frac{1}{2} \log_{10} x\)

- \(\log_{1000} x = \log_{10^3} x = \frac{1}{3} \log_{10} x\)

Substitute these into the equation:

\(4 \log_{10} x + 4 \left(\frac{1}{2} \log_{10} x\right) + 8 \left(\frac{1}{3} \log_{10} x\right) = 13\)

Simplify the equation:

- \(4 \log_{10} x + 2 \log_{10} x + \frac{8}{3} \log_{10} x = 13\)

Add the logarithmic terms:

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

Calculate the coefficients:

\(\frac{12}{3} + \frac{6}{3} + \frac{8}{3} = \frac{26}{3}\)

The equation becomes:

\(\frac{26}{3} \log_{10} x = 13\)

Multiply both sides by \(\frac{3}{26}\):

\(\log_{10} x = \frac{13 \cdot 3}{26} = \frac{39}{26} = \frac{3}{2}\)

Convert the logarithmic form to exponential form to solve for \( x \):

\(x = 10^{\frac{3}{2}}\)

This simplifies to:

\(x = \sqrt{10^3} = \sqrt{1000} = 10 \sqrt{10}\)

Approximating \(\sqrt{10} \approx 3.162\), we find:

\(x \approx 10 \times 3.162 = 31.62\)

Thus, the greatest integer not exceeding \( x \) is:

\(\boxed{31}\)

Verify the computed value falls within the provided range (31, 31). The computed value matches exactly, confirming its correctness. Hence, the greatest integer not exceeding \( x \) is indeed 31.

Answer 2

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.