What is the value of the common logarithm \[\log \left( {10,000} \right)\] \[?\]
Answer
540.6k+ views
Hint: First we have to know the common \[\log \] is the Logarithms in base \[10\] is the power of \[10\] that produces that number. First the given number is expressed as the \[n\] power of \[10\] . Then \[n\] is the value of the common logarithm of that number.
Complete step-by-step answer:
Suppose \[p\] and \[q\] are any two non-zero positive real numbers the following formulas holds:
\[{\log _n}\left( {pq} \right) = {\log _n}\left( p \right) + {\log _n}\left( q \right)\] .
\[{\log _n}\left( {\dfrac{p}{q}} \right) = {\log _n}\left( p \right) - {\log _n}\left( q \right)\] .
\[{\log _n}\left( {{p^a}} \right) = a{\log _n}\left( p \right)\] .
\[{\log _p}\left( q \right) = \dfrac{{{{\log }_n}\left( q \right)}}{{{{\log }_n}\left( p \right)}}\] .
Given \[{\log _{10}}\left( {10000} \right)\] ---(1)
Since we know that \[10000 = {10^4}\] , then (1) becomes
\[{\log _{10}}\left( {{{10}^4}} \right)\]
Using the third formula we get
\[{\log _{10}}\left( {{{10}^4}} \right) = 4{\log _{10}}\left( {10} \right)\]
Using the fourth formula, we get
\[4{\log _{10}}\left( {10} \right) = 4 \times \dfrac{{{{\log }_n}10}}{{{{\log }_n}10}} = 4\]
Hence \[{\log _{10}}\left( {10000} \right) = 4\] .
The domain of the common log as well as the logarithm in any base, is \[x > 0\] . We cannot take a \[\log \] of a negative number, since any positive base cannot produce a negative number, no matter what the power.
So, the correct answer is “4”.
Note: Note that the most common error is forgetting that the function does not exist for values of \[x \leqslant 0\] . The result of the common \[\log \] function is simply the variable \[y\] for the equation \[x = {10^y}\] . As there is no value for \[y\] (in the domain of real numbers) for which \[x \leqslant 0\] , the domain for the inverse function (our common log) is \[0 < x < \infty \] . Also note that if \[{\log _n}y\] is greater than \[{\log _n}x\] . that means that \[y\] is greater than \[x\] by a factor of \[n\] .
Complete step-by-step answer:
Suppose \[p\] and \[q\] are any two non-zero positive real numbers the following formulas holds:
\[{\log _n}\left( {pq} \right) = {\log _n}\left( p \right) + {\log _n}\left( q \right)\] .
\[{\log _n}\left( {\dfrac{p}{q}} \right) = {\log _n}\left( p \right) - {\log _n}\left( q \right)\] .
\[{\log _n}\left( {{p^a}} \right) = a{\log _n}\left( p \right)\] .
\[{\log _p}\left( q \right) = \dfrac{{{{\log }_n}\left( q \right)}}{{{{\log }_n}\left( p \right)}}\] .
Given \[{\log _{10}}\left( {10000} \right)\] ---(1)
Since we know that \[10000 = {10^4}\] , then (1) becomes
\[{\log _{10}}\left( {{{10}^4}} \right)\]
Using the third formula we get
\[{\log _{10}}\left( {{{10}^4}} \right) = 4{\log _{10}}\left( {10} \right)\]
Using the fourth formula, we get
\[4{\log _{10}}\left( {10} \right) = 4 \times \dfrac{{{{\log }_n}10}}{{{{\log }_n}10}} = 4\]
Hence \[{\log _{10}}\left( {10000} \right) = 4\] .
The domain of the common log as well as the logarithm in any base, is \[x > 0\] . We cannot take a \[\log \] of a negative number, since any positive base cannot produce a negative number, no matter what the power.
So, the correct answer is “4”.
Note: Note that the most common error is forgetting that the function does not exist for values of \[x \leqslant 0\] . The result of the common \[\log \] function is simply the variable \[y\] for the equation \[x = {10^y}\] . As there is no value for \[y\] (in the domain of real numbers) for which \[x \leqslant 0\] , the domain for the inverse function (our common log) is \[0 < x < \infty \] . Also note that if \[{\log _n}y\] is greater than \[{\log _n}x\] . that means that \[y\] is greater than \[x\] by a factor of \[n\] .
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