Question

How do you use the change of base formula and a calculator to evaluate ${{\log }_{6}}14$?

Answer 1

Hint: We first try to change the base for the logarithm ${{\log }_{6}}14$ to its exponential form. We sue the formula for ${{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}$. We place the values $b=e,x=14,a=6$ and put the values of the logarithms using a calculator.

Complete step-by-step answer:
We can change the base of the logarithms to the known bases of exponential which can be calculated using the calculator.
Let an arbitrary logarithmic function be $A={{\log }_{a}}x$. The conditions for the expression to be logical is $a,x>0;a\ne 1$.
We have $\log {{x}^{b}}=b\log x$. The power value of $b$ goes as a multiplication with $\log x$.
We also use the formula of logarithm where ${{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}$. For our problem we will assume the value $b=e$. We also know that ${{\log }_{e}}x=\ln x$.
Now we place the values $b=e,x=14,a=6$ in the equation ${{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}$.
We get ${{\log }_{6}}14=\dfrac{{{\log }_{e}}14}{{{\log }_{e}}6}=\dfrac{\ln 14}{\ln 6}$.
Now using the calculator, we find the values for $\ln 14;\ln 6$.
We have $\ln 14=2.639;\ln 6=1.791$.
Putting the values, we get ${{\log }_{6}}14=\dfrac{\ln 14}{\ln 6}=\dfrac{2.639}{1.791}=1.4728$.

Note: There are some particular rules that we follow in case of finding the condensed form of logarithm. We identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power. Sometimes we also use 10 instead of $e$.