Given that $l o g_{10} 5 = 0.70$ and $l o g_{10} 3 = 0.48$ , the value of $l o g_{30} 8$ is

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Solution

$l o g_{10} 5 = 0.70 a n d l o g_{10} 3 = 0.48 l o g_{10} 8 = \frac{l o g_{10} 8}{l o g_{10} 30} = \frac{3 l o g_{10} 2}{l o g_{10} 3 + l o g_{10} 10} = \frac{3 l o g_{10} 2}{0.48 + 1} ∴ l o g_{30} 8 = \frac{3 l o g_{10} 2}{1.48} . . . (1) l o g_{10} 3 = 0.70 \Rightarrow \frac{l o g_{2} 5}{l o g_{2} (5 + 1)} = 0.70 \frac{l o g_{2} 5}{l o g_{2} 3 + l o g_{2} 2} = 0.70 \Rightarrow \frac{l o g_{2} 5}{l o g_{2} (5 + 1)} = 0.70$
Solving for $l o g_{2} 5 w e g e t, l o g_{2} 5 = \frac{7}{3} . . . (2) l o g_{10} 2 = \frac{l o g_{2} 2}{l o g_{2} 10} = \frac{1}{l o g_{2} 2 + l o g_{2} 5} = \frac{1}{1 + l o g_{2} 5} ∴ l o g_{10} 2 = \frac{l o g_{2} 2}{l o g_{2} 10} = \frac{1}{l o g_{2} 2 + l o g_{2} 5} = \frac{1}{1 + l o g_{2} 5} ∴ l o g_{10} 2 = \frac{1}{1 + \frac{7}{3}} = \frac{3}{10} = 0.3 l o g_{10} 2 = 0.3... (4)$
Substitute (4) in (1),
$l o g_{30} 8 = \frac{3 \times 0.3}{1.48} = \frac{0.9}{1.48} ∴ l o g_{30} 8 = 0.61$ .

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