Question

How do you simplify $24\log X-6\log Y$?

Answer 1

We solve the given equation by using the different identity formulas of logarithm like $\ln a-\ln b=\ln \dfrac{a}{b}$, ${{\log }_{e}}a=y\Rightarrow a={{e}^{y}}$. The main step would be to form one single logarithm function instead of two. We solve the linear equation with the help of basic binary operations.

Complete step-by-step solution:
We take the logarithmic identity for the given equation $24\log X-6\log Y$ to find the simplified form.
We have $p{{\log }_{x}}a={{\log }_{x}}{{a}^{p}}$. The subtraction for logarithm works as $\ln a-\ln b=\ln \dfrac{a}{b}$.
We operate the identity $p{{\log }_{x}}a={{\log }_{x}}{{a}^{p}}$ on both parts of the equation $24\log X-6\log Y$.
For $24\log X$, the representations are $p=24,a=X$. So, $24\log X=\log {{X}^{24}}$.
For $6\log Y$, the representations are $p=6,a=Y$. So, $6\log Y=\log {{Y}^{6}}$.
So, $24\log X-6\log Y=\log {{X}^{24}}-\log {{Y}^{6}}$
We operate the subtraction part in $\log {{X}^{24}}-\log {{Y}^{6}}$.
$\log {{X}^{24}}-\log {{Y}^{6}}=\log \dfrac{{{X}^{24}}}{{{Y}^{6}}}$
Therefore, the simplified form of the equation $24\log X-6\log Y$ is $\log \dfrac{{{X}^{24}}}{{{Y}^{6}}}$.

Note: In case of the base is not mentioned then the general solution for the base for logarithm is 10. But the base of $e$ is fixed for $\ln$. We also need to remember that for logarithm function there has to be a domain constraint.
For any ${{\log }_{e}}a$, $a>0$. This means for $24\log X-6\log Y$, $X,Y>0$.
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.