Log Functions

[mathxyz]

I have two questions below.

Write each expression as a sum and/or difference of logs. Express powers as factors.

1) LN [x^(2) - x -2/(x+4)^(2) ]^(1/3) =

My work:

LN = log_e (x^2 - x - 2)^1/3 - log_e (x+4) ^1/3(2)

1/3log_e (x^2 - x - 2) - log_e (x+4) ^2/3

Here's my wrong answer:

1/3log_e(x^2-x-2) - 2/3log_e(x+4)

Where did I go wrong?

I don't know what to do with question 2. Here it is:

Write each expression as a sum and/or difference of logs. Express powers as factors.


2) LN 5x(sqrt{ 1-3x})/(x-4)^(3) =

 

[Matt]

Hi mathxyz,

I have two questions below.

Write each expression as a sum and/or difference of logs. Express powers as factors.

1) LN [x^(2) - x -2/(x+4)^(2) ]^(1/3) =

My work:

LN = log_e (x^2 - x - 2)^1/3 - log_e (x+4) ^1/3(2)

1/3log_e (x^2 - x - 2) - log_e (x+4) ^2/3

Here's my wrong answer:

1/3log_e(x^2-x-2) - 2/3log_e(x+4)

Where did I go wrong?

Good work so far. Why do you think your answer is wrong? It looks fine to me. Perhaps you could try to simplify it further? You might have noticed that you can factor the first polynomial:

. . . . x² - x - 2 = (x - 2)(x + 1)

With this you can expand the first logarithm in your final answer.

 

[Matt]

Hi mathxyz,

I don't know what to do with question 2. Here it is:

Write each expression as a sum and/or difference of logs. Express powers as factors.


2) LN 5x(sqrt{ 1-3x})/(x-4)^(3) =

Another way to write the square root function is with exponents. You might remember from an algebra lesson long ago that:

. . . . √(1 - 3x) = (1 - 3x)^1/2

From your work in the first problem it looks like you have a pretty good handle on logarithms. Do you think you can finish it from here?

 

[mathxyz]

Yes..

I can take it from here.