Question

Find the value of \[{{\log }_{10}}\dfrac{76}{3.8}\]

Answer 1

Hint: We use the formula \[{{\log }_{b}}(mn)={{\log }_{b}}m+{{\log }_{b}}n\] to solve this question. We will begin with converting the decimal in the denominator into a simpler number and then dividing it to get a smaller number and after this we will use the above formula to arrive at the answer.

Complete Step-by-Step solution:
Before proceeding with the question, we must know the definition of logarithm and concepts related to it. A logarithm is defined as the power to which number must be raised to get some other values. It is the most convenient way to express large numbers.
The important rule is \[{{\log }_{b}}(mn)={{\log }_{b}}m+{{\log }_{b}}n......(1)\]. In this rule, the multiplication of two logarithmic values is equal to the addition of their individual logarithms.
So the expression mentioned in the question is,
\[\Rightarrow {{\log }_{10}}\dfrac{76}{3.8}..........(2)\]
Simplifying equation (1) by converting the decimal in denominator into number, we get,
\[\Rightarrow {{\log }_{10}}\dfrac{760}{38}..........(3)\]
Now we know that 760 is divisible by 38. So using this information in equation (2) we get,
\[\Rightarrow {{\log }_{10}}20..........(4)\]
Now we know that 20 is equal to 2 multiplied by 10. So using this information in equation (3) we get,
\[\Rightarrow {{\log }_{10}}(2\times 10)..........(5)\]
Now applying the formula from equation (1) in equation (5) we get,
\[\Rightarrow {{\log }_{10}}2+{{\log }_{10}}10..........(6)\]
Now we know that \[{{\log }_{10}}10=1\], so applying this in equation (6) we get,
\[\Rightarrow {{\log }_{10}}2+1..........(7)\]
Now we know that \[{{\log }_{10}}2=0.30\], so substituting this in equation (7) we get,
\[\Rightarrow 0.30+1=1.30\]
Hence the value of \[{{\log }_{10}}\dfrac{76}{3.8}\] is 1.30.

Note: We have to be thorough with the basic rules of logarithm and apply it to break any larger terms into simpler terms. We may get confused in equation (7) as how to proceed further but we have to remember log values of initial numbers and if we don’t remember then in a hurry we might end up with an incomplete solution.