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Hint: We will write $\log 0.0125$ as $\log \dfrac{125}{10000}$. Now, we know that $\log \left( \dfrac{a}{b} \right)=\log \left( a \right)-\log \left( b \right)$. Therefore $\log \dfrac{125}{10000}$ can be written as $\log 125-\log 10000$. So, to find the value of $\log 0.0125$, we will solve the expression or modified equation $\log 125-\log 10000$.

__Complete step by step answer:__

It is given in the question that we have to find the value of $\log 0.0125$. We are given that the value of $\log 2=0.3010300$, $\log 3=0.4771213$ and $\log 7=0.8450980$. We can write $\log 0.0abc$ as $\log \left( \dfrac{abc}{10000} \right)$. Therefore $\log 0.0125$ can be written as $\log \dfrac{125}{10000}$ as the new modified form.

We also know that $\log \left( \dfrac{a}{b} \right)=\log \left( a \right)-\log \left( b \right)$ is an another relation in the logarithm algebra, therefore we can further modify equation as follows $\log \left( \dfrac{125}{10000} \right)=\log 125-\log 10000$.

Now, $\log 125$ can be written as $\log \left( {{\left( 5 \right)}^{3}} \right)$ and $\log 10000$ can be written as $\log \left( {{\left( 10 \right)}^{4}} \right)$ therefore, equation modified as $\log \left( \dfrac{125}{10000} \right)=\log \left( {{\left( 5 \right)}^{3}} \right)-\log \left( {{\left( 10 \right)}^{4}} \right)$. Now, using the formula $\log \left( {{\left( a \right)}^{b}} \right)=b\times \log a$, we get the equation modified to $\log \left( \dfrac{125}{10000} \right)=3\times \log 5-4\times \log 10$.

The value of $\log 10$ is 1, and $\log 5$ can be written as $\log \left( \dfrac{10}{2} \right)$, therefore putting these value in the above equation, we get \[\log \left( \dfrac{125}{10000} \right)=3\times \log \left( \dfrac{10}{2} \right)-4\times 1\].

Again using the formula $\log \left( \dfrac{a}{b} \right)=\log \left( a \right)-\log \left( b \right)$, $\log \left( \dfrac{10}{2} \right)$ can be written as $\log 10-\log 2$, putting this value, we get, \[\log \left( \dfrac{125}{10000} \right)=3\times \left( \log 10-\log 2 \right)-4\],

therefore solving further, \[\log \left( \dfrac{125}{10000} \right)=3\times 1-3\times \log 2-4\]

\[=\log \left( \dfrac{125}{10000} \right)=3-4-3\times \log 2\]

\[=\log \left( \dfrac{125}{10000} \right)=-1-3\times 0.3010300\]

\[=\log \left( \dfrac{125}{10000} \right)=-1.903\]

Therefore the value of $\log 0.0125$is found out to be -1.903.

Note: Students may get confused by seeing the value of $\log 0.0125$ as -1.903 but it is correct that the negative sign is because of the smaller argument value of log, that is less than 1. Generally, we find log keeping base 10. So, we have used base as 10 and not as e, called as natural logarithm. If we use natural logarithm, we will get a different value as our answer. Conversion from natural logarithm to base 10 logarithm can be done as follows ${{\log }_{10}}\left( x \right)=2.303\times {{\log }_{e}}\left( x \right)$.

It is given in the question that we have to find the value of $\log 0.0125$. We are given that the value of $\log 2=0.3010300$, $\log 3=0.4771213$ and $\log 7=0.8450980$. We can write $\log 0.0abc$ as $\log \left( \dfrac{abc}{10000} \right)$. Therefore $\log 0.0125$ can be written as $\log \dfrac{125}{10000}$ as the new modified form.

We also know that $\log \left( \dfrac{a}{b} \right)=\log \left( a \right)-\log \left( b \right)$ is an another relation in the logarithm algebra, therefore we can further modify equation as follows $\log \left( \dfrac{125}{10000} \right)=\log 125-\log 10000$.

Now, $\log 125$ can be written as $\log \left( {{\left( 5 \right)}^{3}} \right)$ and $\log 10000$ can be written as $\log \left( {{\left( 10 \right)}^{4}} \right)$ therefore, equation modified as $\log \left( \dfrac{125}{10000} \right)=\log \left( {{\left( 5 \right)}^{3}} \right)-\log \left( {{\left( 10 \right)}^{4}} \right)$. Now, using the formula $\log \left( {{\left( a \right)}^{b}} \right)=b\times \log a$, we get the equation modified to $\log \left( \dfrac{125}{10000} \right)=3\times \log 5-4\times \log 10$.

The value of $\log 10$ is 1, and $\log 5$ can be written as $\log \left( \dfrac{10}{2} \right)$, therefore putting these value in the above equation, we get \[\log \left( \dfrac{125}{10000} \right)=3\times \log \left( \dfrac{10}{2} \right)-4\times 1\].

Again using the formula $\log \left( \dfrac{a}{b} \right)=\log \left( a \right)-\log \left( b \right)$, $\log \left( \dfrac{10}{2} \right)$ can be written as $\log 10-\log 2$, putting this value, we get, \[\log \left( \dfrac{125}{10000} \right)=3\times \left( \log 10-\log 2 \right)-4\],

therefore solving further, \[\log \left( \dfrac{125}{10000} \right)=3\times 1-3\times \log 2-4\]

\[=\log \left( \dfrac{125}{10000} \right)=3-4-3\times \log 2\]

\[=\log \left( \dfrac{125}{10000} \right)=-1-3\times 0.3010300\]

\[=\log \left( \dfrac{125}{10000} \right)=-1.903\]

Therefore the value of $\log 0.0125$is found out to be -1.903.

Note: Students may get confused by seeing the value of $\log 0.0125$ as -1.903 but it is correct that the negative sign is because of the smaller argument value of log, that is less than 1. Generally, we find log keeping base 10. So, we have used base as 10 and not as e, called as natural logarithm. If we use natural logarithm, we will get a different value as our answer. Conversion from natural logarithm to base 10 logarithm can be done as follows ${{\log }_{10}}\left( x \right)=2.303\times {{\log }_{e}}\left( x \right)$.