Why is 2/(x lnx^2) = 1 / (x ln x) ?

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In summary, the student is struggling with simplifying the equation 2/(x lnx^2) = 1 / (x ln x) and is unsure of how to proceed. They have looked for information on the "ln simplify process" in their precalculus book but have not found it. They are seeking advice and clarification on the properties of logarithms to help them solve the problem.
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Knight226
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Homework Statement


2/(x lnx^2) = 1 / (x ln x)
That is the answer in the back of my textbook. I have reached 2 ( x ln x^2) , but I do not know why it is = to 1 / ( x ln x )

Homework Equations





The Attempt at a Solution


I tried to look into precal book to see if I could find information on this "ln simplify process" but I can't find it. I didn't have proper math classes so I am missing a lot of math stuff that people who take calculus should know :/ Please advice. Thank you.
 
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  • #2
Knight226 said:

Homework Statement


2/(x lnx^2) = 1 / (x ln x)
That is the answer in the back of my textbook. I have reached 2 ( x ln x^2) , but I do not know why it is = to 1 / ( x ln x )

Homework Equations





The Attempt at a Solution


I tried to look into precal book to see if I could find information on this "ln simplify process" but I can't find it. I didn't have proper math classes so I am missing a lot of math stuff that people who take calculus should know :/ Please advice. Thank you.

Can you simplify ln(x2)? Look in your book for properties of logarithms. They should be close to the beginning of the section that introduces logs.

Also, this section of the forum is for Calculus & Beyond. Yours is a precalculus question.
 

Related to Why is 2/(x lnx^2) = 1 / (x ln x) ?

1. Why does 2/(x lnx^2) equal 1 / (x ln x)?

This equation follows from the properties of logarithms and the rules of algebra. Specifically, the rule for simplifying logarithmic expressions states that loga(x^n) = n * loga(x). In this case, we have logx(x^2) = 2 and logx(x) = 1. Therefore, 2/(x lnx^2) = 2/(x*2) = 1/x = 1 / (x ln x).

2. Can you prove that 2/(x lnx^2) is equal to 1 / (x ln x)?

Yes, we can prove this using the same properties of logarithms mentioned before. First, we can rewrite 2/(x lnx^2) as 2/(x*lnx)^2. Then, using the rule for simplifying logarithmic expressions, we can write this as 2/(x*lnx*lnx) = 2/(x*lnx) * 1/lnx. Finally, we can substitute 1/lnx for logx(x) and cancel out the x's to get 1 / (x ln x).

3. Is there a specific context in which this equation is commonly used?

This equation is commonly used in calculus and physics when dealing with logarithmic functions and their derivatives. It can also be used in solving various mathematical equations and problems involving logarithms and exponential functions.

4. Is there an intuitive explanation for why 2/(x lnx^2) is equal to 1 / (x ln x)?

One way to think about this is by considering the inverse relationship between logarithms and exponential functions. The expression 2/(x lnx^2) can be rewritten as 2/(lnx * (lnx)^2), which is essentially the derivative of (lnx)^2. Similarly, 1 / (x ln x) can be rewritten as (lnx)^-1, which is the inverse of the derivative of lnx. So, in a way, we can say that these two expressions are inverses of each other, resulting in them being equal.

5. Are there any exceptions to this equation?

No, this equation holds true for all values of x except for x = 1, where the expressions on both sides become undefined.

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