Question

How do you find the exact value of ${{\log }_{4}}{{16}^{1.2}}$?

Answer 1

Hint: We solve the given equation ${{\log }_{4}}{{16}^{1.2}}$ using the particular identity formula of logarithm like \[{{\log }_{{{y}^{b}}}}{{x}^{a}}=\dfrac{a}{b}{{\log }_{y}}x\]. The main step would be to eliminate the power value of the logarithm functions and keep it as a simple logarithm. We solve the linear multiplication with the help of basic binary operations.


Complete step by step answer:
We take the logarithmic identity for the given equation ${{\log }_{4}}{{16}^{1.2}}$ to find the solution for condensation. For a condensed form of logarithm, we apply power property, products of factors and logarithm of a power. For our given equation we are only going to apply the power property.
We have \[{{\log }_{{{y}^{b}}}}{{x}^{a}}=\dfrac{a}{b}{{\log }_{y}}x\]. The power value of $a$ goes as a multiplication and $b$ as division with $\log x$.
In case of logarithmic numbers having powers, we have to multiply the power in front of the logarithm to get the single logarithmic function.
We can write $4={{2}^{2}};{{16}^{1.2}}={{\left( {{2}^{4}} \right)}^{1.2}}={{2}^{4.8}}$. Therefore, \[{{\log }_{4}}{{16}^{1.2}}={{\log }_{{{2}^{2}}}}{{2}^{4.8}}\].
We sue the formula of \[{{\log }_{{{y}^{b}}}}{{x}^{a}}=\dfrac{a}{b}{{\log }_{y}}x\] to get
\[{{\log }_{{{2}^{2}}}}{{2}^{4.8}}=\dfrac{4.8}{2}{{\log }_{2}}2=2.4{{\log }_{2}}2\]
We have the identity formula of ${{\log }_{x}}x=1$. This gives ${{\log }_{2}}2=1$.
Putting the value, we get \[2.4{{\log }_{2}}2=2.4\]
Therefore, the simplified form of ${{\log }_{4}}{{16}^{1.2}}$ is \[2.4\].

Note: There are some particular rules that we follow in case of finding the condensed form of logarithm. We first apply the power property first. Then we identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power. Then we apply the product property. Rewrite sums of logarithms as the logarithm of a product. We also have the quotient property rules.