Properties of logarithms?

Question

Use the properties of logarithms to simplify each expression below.

a) ln x(x-2)+(x^2+1)

I believe the correct answer is: ln(x-2)+ln(x^2+1)

b) log4 x^2/(x-1)^3

Answer: 1/2log4

c) ln(x^2 - 2x+1)

Answer: ln(x^2+1)

I am unsure if the answers are right or not. Could you please help me know if I am on the right track or not! :] If not please help!

10 points today! :]

Answer 1

a) This one doesn't make much sense. Be sure to use parentheses properly. It seems like you mean: ln [x(x - 2) + (x² + 1)]. But even that doesn't look right. Are you sure it's not ln [x(x - 2)(x² + 1)] ?

b) By "log4" do you mean "log₄" or "log 4"? I'll assume log₄ and write it simply as "log".

log [x²/(x - 1)³]

log x² - log (x - 1)³

2 log x - 3 log (x - 1)

c) ln (x² - 2x + 1)

ln (x - 1)²

2 ln (x - 1)

Answer 2

a) First, I'm not certain what you mean here;

Do you mean ln [ x(x-2) + (x^2 + 1) ]   or is it

( ln[ x(x-2) ])  + (x^2 + 1 )  ?  You need to use parentheses to make these clear. I'll answer both:

If you mean  ln [ x(x-2) + (x^2 + 1) ]   then first work inside the parenthese to get  ln( x^2 - 2x + x^2 +1 ) = ln( 2x^2 -2x + 1).  This can't be simplified further. NOTE: It is NOT true that

  ln(a+b) = ( ln a )+ ( ln b ).

If you mean  ( ln[ x(x-2) ])  + (x^2 + 1 ) , then we get for the first term:

ln[ x(x-2) ] = ln(x) + ln(x-2)   {because ln(ab) = ln a + ln b }.

So ( ln[ x(x-2) ])  + (x^2 + 1 )  would = ln(x) + ln(x-2)  + (x^2 + 1).

This can't be simplified further.

b) I assume you mean log base 4 of  x^2 / (x-1)^3. (In printed books, log base 4 would be printed as log with a small number 4 subscript after the g).

Then first use the rule that  log(a/b) = log a - log b to get:

log [x^2 / (x-1)^3]  =  log(x^2) - log[(x-1)^3]

Then use the rule that  log a^b = b*log a  to get:

= 2*log(x) - 3*log(x-1).

These rules of log are true in any base log. If the question started in log base 4, then all the logs afterward in the answer are also in base 4.

c)  ln(x^2 - 2x + 1)  =  ln[ (x-1)^2 ] = 2*ln(x-1)

Again, note that ln(a+b) does NOT equal ln a + ln b

I don't see how you got the answer ln(x^2 +1). It is clearly wrong. You can illustrate it is wrong by letting x = 2.

Then ln(x^2-2x+1) = ln(1), but

ln(x^2 + 1) = ln(5) and ln(1) does not equal ln(5).

Let me know if you need more details here or if I'm misunderstanding what's written.