Questions & Answers

Question

Answers

${{\log }_{10}}\left( \dfrac{12}{5} \right)+{{\log }_{10}}\left( \dfrac{25}{21} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right)$

Answer

Verified

112.8K+ Views

Hint: Use the basic properties of logarithmic functions, which are given as

$\begin{align}

& {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( mn \right), \\

& {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right) \\

\end{align}$

And the value of $'\log '$ of a number with the same base is ‘1’ or mathematically we can write the value of ${{\log }_{a}}a$ as 1. Use these concepts to solve the given expression.

Complete step-by-step answer:

As we know the property of logarithm function of adding two logarithm function and subtracting two logarithm function on the same base are given as

$\begin{align}

& {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}mn.............\left( i \right) \\

& {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right).............\left( ii \right) \\

\end{align}$

Now, coming to the question we have the equation as

${{\log }_{10}}\left( \dfrac{12}{5} \right)+{{\log }_{10}}\left( \dfrac{25}{21} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right)$

Let the value of the expression be ‘s’ so, we can write value of the given expression as

$s={{\log }_{10}}\left( \dfrac{12}{5} \right)+{{\log }_{10}}\left( \dfrac{25}{21} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right).................\left( iii \right)$

Now, we can observe that the first two terms of the relation are in summation from and have the same base to both of them. It means we can apply the identity of equation (i) with the first two terms of the equation (iii). So, we get

$\begin{align}

& s={{\log }_{10}}\left( \dfrac{12}{5}\times \dfrac{25}{21} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right), \\

& s={{\log }_{10}}\left( \dfrac{4}{1}\times \dfrac{5}{7} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right), \\

& s={{\log }_{10}}\left( \dfrac{20}{7} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right)..................\left( iv \right) \\

\end{align}$

Now, we can observe that the two terms of expression (iv) are in different forms and have the same base as well. It means we can apply equation (ii) with the terms of the equation (iv). Hence, we get

\[\begin{align}

& s={{\log }_{10}}\left( \dfrac{\left( \dfrac{20}{7} \right)}{\left( \dfrac{2}{7} \right)} \right)={{\log }_{10}}\left( \dfrac{20}{7}\times \dfrac{7}{2} \right) \\

& s={{\log }_{10}}10....................(v) \\

\end{align}\]

Now, we know the property of logarithm functions that any logarithm function with same value and same base will give value as 1 i.e. value of the expression ${{\log }_{a}}a=1.$ Therefore, value of the given expression in the problem i.e. ${{\log }_{10}}\left( \dfrac{12}{5} \right)+{{\log }_{10}}\left( \dfrac{25}{21} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right)$ is 1.

Note: One may solve the given expression in one step as well. We can apply the properties

$\begin{align}

& {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}mn, \\

& {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right) \\

\end{align}$

In one step. So, we get value of expression as

${{\log }_{10}}\left( \dfrac{\dfrac{12}{5}\times \dfrac{25}{21}}{\left( \dfrac{2}{7} \right)} \right)={{\log }_{10}}10=1$

Don’t confuse the properties of the logarithm functions. As one may apply identities as

$\begin{align}

& {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right), \\

& {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}mn \\

\end{align}$

Which are wrong and just opposite to one another. So, use the correct results of the properties and apply them carefully as well. Another approach for the problem would be that we can use same properties mentioned above in reverse manner i.e. use the same properties as

$\begin{align}

& {{\log }_{a}}mn={{\log }_{a}}m+{{\log }_{a}}n, \\

& {{\log }_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n. \\

\end{align}$

So, one may write the given expression as

$\begin{align}

& {{\log }_{10}}12-{{\log }_{10}}5+{{\log }_{10}}25-{{\log }_{10}}21-\left( {{\log }_{10}}2-{{\log }_{10}}7 \right) \\

& ={{\log }_{10}}3+{{\log }_{10}}4-{{\log }_{10}}5+2{{\log }_{10}}5-{{\log }_{10}}7-{{\log }_{10}}3-{{\log }_{10}}2+{{\log }_{10}}7 \\

\end{align}$

Now, simplifying them we get

${{\log }_{10}}2+{{\log }_{10}}5={{\log }_{10}}10=1$

Where, we need to split the logarithm functions in simplest form using the properties mentioned in the properties above.

$\begin{align}

& {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( mn \right), \\

& {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right) \\

\end{align}$

And the value of $'\log '$ of a number with the same base is ‘1’ or mathematically we can write the value of ${{\log }_{a}}a$ as 1. Use these concepts to solve the given expression.

Complete step-by-step answer:

As we know the property of logarithm function of adding two logarithm function and subtracting two logarithm function on the same base are given as

$\begin{align}

& {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}mn.............\left( i \right) \\

& {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right).............\left( ii \right) \\

\end{align}$

Now, coming to the question we have the equation as

${{\log }_{10}}\left( \dfrac{12}{5} \right)+{{\log }_{10}}\left( \dfrac{25}{21} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right)$

Let the value of the expression be ‘s’ so, we can write value of the given expression as

$s={{\log }_{10}}\left( \dfrac{12}{5} \right)+{{\log }_{10}}\left( \dfrac{25}{21} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right).................\left( iii \right)$

Now, we can observe that the first two terms of the relation are in summation from and have the same base to both of them. It means we can apply the identity of equation (i) with the first two terms of the equation (iii). So, we get

$\begin{align}

& s={{\log }_{10}}\left( \dfrac{12}{5}\times \dfrac{25}{21} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right), \\

& s={{\log }_{10}}\left( \dfrac{4}{1}\times \dfrac{5}{7} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right), \\

& s={{\log }_{10}}\left( \dfrac{20}{7} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right)..................\left( iv \right) \\

\end{align}$

Now, we can observe that the two terms of expression (iv) are in different forms and have the same base as well. It means we can apply equation (ii) with the terms of the equation (iv). Hence, we get

\[\begin{align}

& s={{\log }_{10}}\left( \dfrac{\left( \dfrac{20}{7} \right)}{\left( \dfrac{2}{7} \right)} \right)={{\log }_{10}}\left( \dfrac{20}{7}\times \dfrac{7}{2} \right) \\

& s={{\log }_{10}}10....................(v) \\

\end{align}\]

Now, we know the property of logarithm functions that any logarithm function with same value and same base will give value as 1 i.e. value of the expression ${{\log }_{a}}a=1.$ Therefore, value of the given expression in the problem i.e. ${{\log }_{10}}\left( \dfrac{12}{5} \right)+{{\log }_{10}}\left( \dfrac{25}{21} \right)-{{\log }_{10}}\left( \dfrac{2}{7} \right)$ is 1.

Note: One may solve the given expression in one step as well. We can apply the properties

$\begin{align}

& {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}mn, \\

& {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right) \\

\end{align}$

In one step. So, we get value of expression as

${{\log }_{10}}\left( \dfrac{\dfrac{12}{5}\times \dfrac{25}{21}}{\left( \dfrac{2}{7} \right)} \right)={{\log }_{10}}10=1$

Don’t confuse the properties of the logarithm functions. As one may apply identities as

$\begin{align}

& {{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right), \\

& {{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}mn \\

\end{align}$

Which are wrong and just opposite to one another. So, use the correct results of the properties and apply them carefully as well. Another approach for the problem would be that we can use same properties mentioned above in reverse manner i.e. use the same properties as

$\begin{align}

& {{\log }_{a}}mn={{\log }_{a}}m+{{\log }_{a}}n, \\

& {{\log }_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n. \\

\end{align}$

So, one may write the given expression as

$\begin{align}

& {{\log }_{10}}12-{{\log }_{10}}5+{{\log }_{10}}25-{{\log }_{10}}21-\left( {{\log }_{10}}2-{{\log }_{10}}7 \right) \\

& ={{\log }_{10}}3+{{\log }_{10}}4-{{\log }_{10}}5+2{{\log }_{10}}5-{{\log }_{10}}7-{{\log }_{10}}3-{{\log }_{10}}2+{{\log }_{10}}7 \\

\end{align}$

Now, simplifying them we get

${{\log }_{10}}2+{{\log }_{10}}5={{\log }_{10}}10=1$

Where, we need to split the logarithm functions in simplest form using the properties mentioned in the properties above.

Students Also Read