Question

Find the value of ${10^{{{\log }_{10}}\left( {\dfrac{3}{4}} \right)}} \cdot {5^{{{\log }_5}\left( {\dfrac{3}{4}} \right)}} \cdot {3^{{{\log }_3}7}}$.

Answer 1

Here, we will apply the rules of logarithms to each term of the given expression separately. Then we will substitute the value of each term in the given expression and simplify it further. Then, we will multiply the terms to find the required value.

Formula used: We will use the rule of logarithm ${b^{{{\log }_b}x}} = x$.

Complete step-by-step answer:
We will use the rules of logarithms to simplify the expression.
Rule of logarithm: ${b^{{{\log }_b}x}} = x$ where $b > 0$, $b \ne 1$, and $x$ is a real number.
First, we will simplify the expression ${10^{{{\log }_{10}}\left( {\dfrac{3}{4}} \right)}}$.
We can observe that the expression is of the form ${b^{{{\log }_b}x}}$.
Substituting $b = 10$ and $x = \dfrac{3}{4}$ in the rule of logarithm ${b^{{{\log }_b}x}} = x$, we get
${10^{{{\log }_{10}}\left( {\dfrac{3}{4}} \right)}} = \dfrac{3}{4}$
Next, we will simplify the expression ${5^{{{\log }_5}\left( {\dfrac{3}{4}} \right)}}$.
We can observe that the expression is of the form ${b^{{{\log }_b}x}}$.
Substituting $b = 5$ and $x = \dfrac{3}{4}$ in the rule of logarithm ${b^{{{\log }_b}x}} = x$, we get
${5^{{{\log }_5}\left( {\dfrac{3}{4}} \right)}} = \dfrac{3}{4}$
Next, we will simplify the expression ${3^{{{\log }_3}7}}$.
We can observe that the expression is of the form ${b^{{{\log }_b}x}}$.
Substituting $b = 3$ and $x = 7$ in the rule of logarithm ${b^{{{\log }_b}x}} = x$, we get
${3^{{{\log }_3}7}} = 7$
Now, we will simplify the given expression.
Substituting ${10^{{{\log }_{10}}\left( {\dfrac{3}{4}} \right)}} = \dfrac{3}{4}$, ${5^{{{\log }_5}\left( {\dfrac{3}{4}} \right)}} = \dfrac{3}{4}$, and ${3^{{{\log }_3}7}} = 7$ in the expression ${10^{{{\log }_{10}}\left( {\dfrac{3}{4}} \right)}} \cdot {5^{{{\log }_5}\left( {\dfrac{3}{4}} \right)}} \cdot {3^{{{\log }_3}7}}$, we get
${10^{{{\log }_{10}}\left( {\dfrac{3}{4}} \right)}} \cdot {5^{{{\log }_5}\left( {\dfrac{3}{4}} \right)}} \cdot {3^{{{\log }_3}7}} = \dfrac{3}{4} \cdot \dfrac{3}{4} \cdot 7$
Multiplying the terms in the expression, we get
$\begin{array}{l} \Rightarrow {10^{{{\log }_{10}}\left( {\dfrac{3}{4}} \right)}} \cdot {5^{{{\log }_5}\left( {\dfrac{3}{4}} \right)}} \cdot {3^{{{\log }_3}7}} = \dfrac{9}{{16}} \cdot 7\\\ \Rightarrow {10^{{{\log }_{10}}\left( {\dfrac{3}{4}} \right)}} \cdot {5^{{{\log }_5}\left( {\dfrac{3}{4}} \right)}} \cdot {3^{{{\log }_3}7}} = \dfrac{{63}}{{16}}\end{array}$
Since 63 and 16 are co-prime numbers, we cannot simplify the fraction further.
Therefore, we get the value of the expression ${10^{{{\log }_{10}}\left( {\dfrac{3}{4}} \right)}} \cdot {5^{{{\log }_5}\left( {\dfrac{3}{4}} \right)}} \cdot {3^{{{\log }_3}7}}$ as $\dfrac{{63}}{{16}}$.

Note: We used the term co-prime numbers in the solution. Two numbers are called co-prime numbers if they do not share a common factor other than 1. For example, the factors of 63 are 1, 3, 7, 9, 21, and 63. The factors of 16 are 1, 2, 4, 8, and 16. We can observe that 63 and 16 do not have any common factor other than 1. Therefore, 63 and 16 are co-prime numbers. Hence, the fraction $\dfrac{{63}}{{16}}$ cannot be simplified further.