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

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:
$\boxed{31}.$