Question

How do you calculate ${{\log }_{6}}5$ with a calculator?

Answer 1

Hint: We have been given a logarithmic function whose base is equal to 6 and the term with logarithm is 5. We shall use various properties of the functions in logarithm to simplify and calculate the given expression. The main property that will be used here is the change of base property of the logarithm functions. We shall then put the appropriate values of the required constant logarithmic functions with the help of a calculator.

Complete step by step solution:
Given that ${{\log }_{6}}5$.
In order to simplify this expression, we shall use the basic property of logarithms namely the change of base property of logarithmic functions. According to this property, the base of the logarithm function can be changed to base e (natural log), but then the function would be expressed as the division of logarithm base e of the argument divided by the logarithm base e of the base given in the original logarithm in the question, that is, ${{\log }_{a}}b=\dfrac{{{\log }_{e}}b}{{{\log }_{e}}a}$.
$\Rightarrow {{\log }_{6}}5=\dfrac{{{\log }_{e}}5}{{{\log }_{e}}6}$
Now, since $6=2\times 3$, the expression can be modified as
$\Rightarrow {{\log }_{6}}5=\dfrac{{{\log }_{e}}5}{{{\log }_{e}}2\times 3}$
Here, we shall use another property according to which when a logarithm function is expressed in terms of the product of n terms, then that logarithm function can be expressed as the sum of n individual logarithmic functions of those n terms, that is, $\log ab=\log a+\log b$.
$\Rightarrow {{\log }_{6}}5=\dfrac{{{\log }_{e}}5}{{{\log }_{e}}2+{{\log }_{e}}3}$
We know that ${{\log }_{e}}2=0.693,{{\log }_{e}}3=1.098$ and ${{\log }_{e}}5=1.609$. Substituting these values, we get
$\Rightarrow {{\log }_{6}}5=\dfrac{1.609}{0.693+1.098}$
$\Rightarrow {{\log }_{6}}5=\dfrac{1.609}{1.791}$
$\Rightarrow {{\log }_{6}}5=0.898$
Therefore, ${{\log }_{6}}5$ with a calculator is calculated equal to 0.898.

Note: We must have prior knowledge of the basic properties of the logarithmic functions. One mistake we could have made here was that we could have changed the base of the logarithm equal to 10 instead of changing the base equal to e.