Expert: Sherry Wallin - 7/19/2009

Question
QUESTION: Hi, sherry wallin, i worked out  this problem can u please
check ?
Write the following into single logarithms.

1/2ln(x^2+1)-(4)ln(1/2) -1/2 [ln(x-4)+lnx)]

i got ln 1/8 (x+1)

can u please check and tell me if im right ?
thank you

ANSWER: Raja~
   Please show me all the steps you used to arrive at this answer. You are off by a lot.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: hi !!

ln(x^2+1)^1/2-ln 1/8 - ln (x-4)^1/2 + ln x^1/2  -->
sent the 1/2 in the first part, the 4 in the second part and
the 1/2 in the last part to the back based loga ^r = r loga M

ln(x^2^1/2 + 1^1/2) - ln 1/8 - ln (x^1/2- 4 ^1/2 ) + 1n
√x--> distributing and the numbers

ln(x+1) - ln 1/8 - ln (√x-2) + ln √x --> solved equation

      1/8
ln    ----
     (x+1)
  ---------- + ln √x  --> used the  log formula loga
(M/N)=
     √x-2       loga M-loga N


      1/8(x+1)
ln   --------- + ln √x
        √x-2

         1/8 (x+1)
ln √x    ----------
         √x-2

ln  -1/8 (x+1)
    ----------
         2


thank you so much for actually looking into my question. I
really appreciate your help. Thanks again.

ANSWER: One way to know if your answer is equivalent to the original is to substitute values into the original expression and into your answer and see if they are the same. Use x = 6 to check it.

Your first mistake is you CANNOT distribute the square root (1/2 power) to a sum!!

1/2ln(x^2+1)-(4)ln(1/2) -1/2 [ln(x-4)+lnx)]
= .5ln(x^2+1)-.5[ln(x-4)+lnx)]-(4)ln(.5)
= ln[(x^2+1)/(x(x-4))]^.5 -ln(.5)^4
= ln([(x^2+1)/(x(x-4))]^.5 /(.5)^4)
= ln([(x^2+1)/(x(x-4))]^.5 /((.5)^8)^.5)
= ln [(x^2+1)/((.5^8)(x(x-4)))]

Math Prof


---------- FOLLOW-UP ----------

QUESTION: hi, this is raja. i am really sorry to bother you, but can you
please  explain the last two steps !

Answer
Sure can Raja~ I will number them 1-3 and explain how I went from 1-2 and 2-3:
    
1)  = ln([(x^2+1)/(x(x-4))]^.5 /(.5)^4)
2)  = ln([(x^2+1)/(x(x-4))]^.5 /((.5)^8)^.5)
3)  = ln [(x^2+1)/((.5^8)(x(x-4)))] **

from step 1-2: I used .5 for (1/2) and in this step you have the log[(a/b)^(1/2))/(1/2)^4] where a = x^2+1 and b = x(x-4). So I took (1/2)^4 and to put (1/2)^4  under the square root I needed to square (((1/2)^4)^2)^(1/2) = [(1/2)^8]^(1/2)]] {4(2)(1/2) = 4 and
8(1/2)= 4}

from step 2-3:  since all factors are to the one-half power you have a complex fraction (a/b/z/1)^(1/2) where z = .5^8. Simplifying the complex fraction you have (a/bz)^(1/2) which becomes
ln[(x^2+1)/(x(x-4)(.5^8))]^(1/2) or
(1/2)ln[([(x^2+1)/(x(x-4))]^.5 /(.5)^4)]


I see that I left the 1/2 power off my answer (**) above, sorry

Math Prof