Completely expand as a sum/difference of logs

  • Thread starter Thread starter lolilovepie
  • Start date Start date
AI Thread Summary
The discussion focuses on expanding the logarithmic expression log[(2x^4(x-15)^3)/(12√(x^4-16))] into a sum and difference of logs. Participants clarify the correct interpretation of the numerator and emphasize the importance of using parentheses for readability. The properties of logarithms, such as log(AB) = log(A) + log(B) and the power rule, are discussed to aid in the expansion process. There are corrections made regarding the use of square roots and exponents, ensuring accurate application of logarithmic rules. The final expression remains incomplete, indicating further steps are needed for full expansion.
lolilovepie
Messages
7
Reaction score
0

Homework Statement



Completely expand as a sum/difference of logs
log [ ( 2x^4(x-15)^3) / (12 sqrt (x^4-16) ]

Homework Equations


The Attempt at a Solution



log 2x^4(x-15)^3 - log 12 sqrt (x^4-16)
3 log 2x^4(x-15) - 1/2log 12 (x^4-16)
 
Last edited:
Physics news on Phys.org
lolilovepie said:

Homework Statement



Completely expand as a sum/difference of logs
log [ ( 2x^4(x-15)^3) / (12 sqrt (x^4-16) ]
Some clarification, please. The numerator of what you wrote is
2x4 * (x - 15)3.

If that isn't what you meant, please use parentheses or brackets to make it clearer.
lolilovepie said:

Homework Equations





The Attempt at a Solution



log 2x^4(x-15)^3 - log 12 sqrt (x^4-16)
3 log 2x^4(x-15) - 1/3log 12 (x^4-16)

Some parentheses would make this more readable, as would the inclusion of = for things that are equal.

Also, sqrt(x) = x1/2, not x1/3.
 
Mark44 said:
Some clarification, please. The numerator of what you wrote is
2x4 * (x - 15)3.

If that isn't what you meant, please use parentheses or brackets to make it clearer.


Some parentheses would make this more readable, as would the inclusion of = for things that are equal.

Also, sqrt(x) = x1/2, not x1/3.

1) yeah, that's what i meant to say, but i didn't know how to do the exponents

2) ok okay thanks! did I solve the problem right because I'm not very sure :\
 
Starting here --
log [ ( 2x^4 * (x-15)^3) / (12 sqrt (x^4-16) ]
= log [ 2x^4 * (x-15)^3] - log[12 * (x^4-16)^(1/2)]
Can you continue?

Notice that I added * to indicate multiplication. I think that's what you intended, but am not sure.

There are some properties of logs that you either don't know or aren't using, such as log(AB) = log(A) + log(B), assuming both A and B are positive.
 
Mark44 said:
Starting here --
log [ ( 2x^4 * (x-15)^3) / (12 sqrt (x^4-16) ]
= log [ 2x^4 * (x-15)^3] - log[12 * (x^4-16)^(1/2)]
Can you continue?

Notice that I added * to indicate multiplication. I think that's what you intended, but am not sure.

There are some properties of logs that you either don't know or aren't using, such as log(AB) = log(A) + log(B), assuming both A and B are positive.

after continuing would it be:

[log 2x^4 + log (x-15)^3] - [log 12 + log (x^4-16)^1/2]

and then use the power rule?

[4 log 2x + 3 log (x-15)] - [log 12 + 1/2 log (x^4-16)]
 
lolilovepie said:
after continuing would it be:

[log 2x^4 + log (x-15)^3] - [log 12 + log (x^4-16)^1/2]

and then use the power rule?

[4 log 2x + 3 log (x-15)] - [log 12 + 1/2 log (x^4-16)]

It's still not completely expanded.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top