Answer
385.5k+ views
Hint: 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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)