Combine into a single logarithm?

Question

Help please! I usually can do these but this is over my head!

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

Answer 1

You need to use three rules to combine these logs:

(a) Subtracting logs is the same as taking the log of the quotient.
ln(a) - ln(b) = ln(a/b)

(b) Adding logs is the same as taking the log of the product.
ln(a) + ln(b) = ln(a*b)

(c) Multiplying a log is the same as taking the log of the exponentition
a*ln(b) = ln(b^a)

First, simplify the individual terms:

(a) 1/2ln(x^2+1) = ln(sqrt(x^2 + 1))
(b) 4ln(1/2) = ln((1/2)^4) = ln(1/16)
(c) 1/2[ln(x-4)+ln x] = 1/2 ln(x*(x-4)) = ln(sqrt(x*(x-4)))

Then combine:

1/2ln(x^2+1) - 4ln(1/2) - 1/2[ln(x-4)+ln x] =
ln(sqrt(x^2 + 1)) - ln(1/16) - ln(sqrt(x*(x-4))) =
ln(sqrt(x^2 + 1) / (1/16) / sqrt(x*(x-4))) =
ln( 16 * sqrt(x^2 + 1) / sqrt(x*(x-4)) ) =
ln( 16 * sqrt( (x^2 + 1) / (x*(x-4)) ) )

PS - Previous respondent had 16 in the denominator, which is incorrect. "- ln(1/16)" is the same as "+ ln(16)"

Answer 2

Combining this is a ton of fun.......arghh. But since it's you. haha

1/2ln(x^2+1) => ln(sqrt(x^2+1))
-4ln(1/2) => ln ((1/2)^-4) => ln(16)
-1/2ln(x-4) => ln((x-4)^(-1/2)) =>ln(1/sqrt(x-4))
-1/2lnx => ln(1/sqrt(x))
so now add/multiply logarithmically
sqrt(x^2+1)*16*1/sqrt(x-4)*1/sqrt(x)
=
16sqrt(x^2+1)
-------------------- <-Division
sqrt(x^2 - 4x)

now add ln in front of above and you have single logarithm.

Now get out of the house for a while............what are you doing logs at this point for anyway............kids these days geez.

Answer 3

ln a + ln b = ln(a*b)
ln a- ln b =ln(a/b)
k*ln a=l(a^k)

1/2ln(x^2+1) - 4ln(1/2) - 1/2[ln(x-4)+ln x]=
ln (x^2+1)^1/2 - ln(1/2)^4 - ln(x-4)^1/2 + ln x^1/2=
ln [(x^2+1)^1/2 * (x^1/2) / [(x-4)^1/2 * 1/16]=
ln {1/16 * [(x^2+1)*x/(x-4)]^1/2}

Answer 4

On #1 and the last part of #3 you left out the operation signs between the logs #2 is ln [ (a + b)(a - b)^5 / c^3 ]