The value of ${log}_{10} 125 - 3 (1 - {log}_{10} 2)$ is equal to

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Solution

The correct option is A $0$
${log}_{10} 125 - 3 (1 - {log}_{10} 2) = x$
${log}_{10} 125 - 3 + {log}_{10} 8 = x$ ...( $log a^{b} = b log a)$
$∴ {log}_{10} 125 - 3 {log}_{10} 10 + {log}_{10} 8 = x$ ...( $log \frac{a}{b} = log a - log b$ )
$∴ x = {log}_{10} (\frac{125 \times 8}{1000}) = {log}_{10} 1 = 0$

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Similar questions

(i) If $l o g (a + 1) = l o g (4 a - 3) - l o g 3$ ; find a.

(ii) If $2 l o g y - l o g x - 3 = 0$ , express x in terms of y.

(iii) Prove that : $l o g_{10} 125 = 3 (1 - l o g_{10} 2)$ .

({log}_{10} 2) + 1

{log}_{10} m

m

α

β

x^{2} + (3)^{1 / 4} x + 3^{1 / 2} = 0,

α^{96} (α^{12} - 1) + β^{96} (β^{12} - 1)

\frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{10^{- \frac{1}{5}} \times 5^{\frac{3}{5}}} \div \frac{4^{- \frac{2}{3}} \times 5^{- \frac{7}{5}}}{4^{- \frac{3}{5}} \times 6^{- \frac{1}{3}}}

The Value of $31^{2}$ is equal to

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