
The logarithm of 0.001 to the base 10 is equal to:
(A) 2
(B) -3
(C) -1
(D) 6
Answer
595.8k+ views
Hint: At first write 0.001 as a fraction. Then apply the following formulas of logarithm :
${{\log }_{x}} {\left( \dfrac{a}{b} \right)}={{\log }_{x}} {a}-{{\log }_{x}} {b},{{\log }_{x}} {{{y}^{n}}}=n{{\log }_{x}} {y},{{\log }_{x}} {x}=1$
Complete step-by-step answer:
We know that the logarithm is the inverse function to exponentiation. That means, if we have:
${{b}^{x}}=y$
$\Rightarrow x={{\log }_{b}} {y}$
The b is known as the base.
In the given question the base is 10. Therefore we have,
${{\log }_{10}} {0.001}$
Now we can write 0.0001 as $\dfrac{1}{1000}$.
${{\log }_{10}} {0.001}={{\log }_{10}} {\left( \dfrac{1}{1000} \right)}$
Now we can apply the following formula:
${{\log }_{x}} {\left( \dfrac{a}{b} \right)}={{\log }_{x}} {a}-{{\log }_{x}} {b}$
Therefore,
${{\log }_{10}} {0.001}={{\log }_{10}} {1}-{{\log }_{10}} {1000}$
Now we know that, ${{\log }_{x}} {1}=0$
Therefore we have,
${{\log }_{10}} {0.001}=0-{{\log }_{10}} {{{10}^{3}}}$
Now we will apply the formula, ${{\log }_{x}} {{{y}^{n}}}=n{{\log }_{x}} {y}$
Therefore,
${{\log }_{10}} {0.001}=-3{{\log }_{10}} {10}$
Now we know that, ${{\log }_{x}} {x}=1$.
Therefore,
${{\log }_{10}} {0.001}=-3\times 1=-3$
Hence, the logarithm of 0.001 to the base 10 is equal to -3.
Therefore, option (b) is correct.
Note: Alternatively we can solve this question by using the exponential form. Let the logarithm of 0.001 to the base 10 is equal to n. That means,
$0.001={{10}^{n}}$
$\Rightarrow \dfrac{1}{1000}={{10}^{n}}$
$\Rightarrow \dfrac{1}{{{10}^{3}}}={{10}^{n}}$
$\Rightarrow {{10}^{-n}}={{10}^{3}}$
$\Rightarrow n=-3$
Hence, option (b) is correct.
${{\log }_{x}} {\left( \dfrac{a}{b} \right)}={{\log }_{x}} {a}-{{\log }_{x}} {b},{{\log }_{x}} {{{y}^{n}}}=n{{\log }_{x}} {y},{{\log }_{x}} {x}=1$
Complete step-by-step answer:
We know that the logarithm is the inverse function to exponentiation. That means, if we have:
${{b}^{x}}=y$
$\Rightarrow x={{\log }_{b}} {y}$
The b is known as the base.
In the given question the base is 10. Therefore we have,
${{\log }_{10}} {0.001}$
Now we can write 0.0001 as $\dfrac{1}{1000}$.
${{\log }_{10}} {0.001}={{\log }_{10}} {\left( \dfrac{1}{1000} \right)}$
Now we can apply the following formula:
${{\log }_{x}} {\left( \dfrac{a}{b} \right)}={{\log }_{x}} {a}-{{\log }_{x}} {b}$
Therefore,
${{\log }_{10}} {0.001}={{\log }_{10}} {1}-{{\log }_{10}} {1000}$
Now we know that, ${{\log }_{x}} {1}=0$
Therefore we have,
${{\log }_{10}} {0.001}=0-{{\log }_{10}} {{{10}^{3}}}$
Now we will apply the formula, ${{\log }_{x}} {{{y}^{n}}}=n{{\log }_{x}} {y}$
Therefore,
${{\log }_{10}} {0.001}=-3{{\log }_{10}} {10}$
Now we know that, ${{\log }_{x}} {x}=1$.
Therefore,
${{\log }_{10}} {0.001}=-3\times 1=-3$
Hence, the logarithm of 0.001 to the base 10 is equal to -3.
Therefore, option (b) is correct.
Note: Alternatively we can solve this question by using the exponential form. Let the logarithm of 0.001 to the base 10 is equal to n. That means,
$0.001={{10}^{n}}$
$\Rightarrow \dfrac{1}{1000}={{10}^{n}}$
$\Rightarrow \dfrac{1}{{{10}^{3}}}={{10}^{n}}$
$\Rightarrow {{10}^{-n}}={{10}^{3}}$
$\Rightarrow n=-3$
Hence, option (b) is correct.
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