Question

X is a +ve real no, 4 log₁₀ (x) + 4log ₁₀₀ (x) + 8 log₁₀₀₀ (x) = 13, then the greatest integer not exceeding 'x'

Answer 1

Correct Answer: 31

We are solving the equation:
\[4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13.\]
Step 1: Rewrite the logs in base 10
Using the properties of logarithms:
\[\log_{100} x = \frac{\log_{10} x}{\log_{10} 100} = \frac{\log_{10} x}{2} = \frac{1}{2} \log_{10} x,\]
\[\log_{1000} x = \frac{\log_{10} x}{\log_{10} 1000} = \frac{\log_{10} x}{3} = \frac{1}{3} \log_{10} x.\]
Substitute these into the equation:
\[4 \log_{10} x + 4 \times \frac{1}{2} \log_{10} x + 8 \times \frac{1}{3} \log_{10} x = 13.\]
Step 2: Simplify the terms
Simplify each term:
\[4 \log_{10} x + 2 \log_{10} x + \frac{8}{3} \log_{10} x = 13.\]
Combine the coefficients:
\[\left( 4 + 2 + \frac{8}{3} \right) \log_{10} x = 13.\]
Find the sum of the coefficients:
\[4 + 2 + \frac{8}{3} = \frac{12}{3} + \frac{6}{3} + \frac{8}{3} = \frac{26}{3}.\]
Thus:
\[\frac{26}{3} \log_{10} x = 13.\]
Step 3: Solve for \(\log_{10} x\)
Multiply through by \(\frac{3}{26}\):
\[\log_{10} x = 13 \times \frac{3}{26} = \frac{39}{26} = \frac{3}{2}.\]
Step 4: Solve for \(x\)
From \(\log_{10} x = \frac{3}{2}\), rewrite in exponential form:
\[x = 10^{\frac{3}{2}} = \sqrt{10^3} = \sqrt{1000}.\]
Simplify:
\[x = 31.622.\]
Step 5: Greatest integer not exceeding \(x\)
The greatest integer not exceeding \(x = 31.622\) is:
\[\boxed{31}.\]