Write the expression as a single logarithm of a single quantity 2lnx-2lnx plus 1 - lnx-1 plus 2ln5?

Answer 1

I am interpreting this as:

2ln(x)-2ln(x+1)-ln(x-1)+2ln(5)

The rules of logarithmic manipulation that will be used for this are:

a*ln(b) = ln(ba)

ln(a)-ln(b)=ln(a/b)

ln(a)+ln(b)=ln(a*b)

Using the first rule, we can perform all of the following operations:

2ln(x) = ln(x2)

2ln(x+1) = ln(x+1)2

2ln(5) = ln(52) = ln(25)

From this we can rewrite the problem thus far as:

ln(x2)-ln((x+1)2)-ln(x-1)+ln(25)

Now using the second and third rules we can do this:

ln(x2) - ln((x+1)2) = ln(x2/(x+1)2)

ln(x2/(x+1)2)-ln(x-1) = ln((x2/(x+1)2)/(x-1)) = ln(x2/((x+1)2(x-1)))

ln(x2/((x+1)2(x-1)))+ln(25) = ln(25x2/((x+1)2(x-1)))

So, we have inside of the natural logarithm function a large fraction whose numerator is 25x2 and whose denominator is (x+1)2(x-1). We can expand the denominator through simple binomial multiplication:

(x+1)2=x2+2x+1

(x2+2x+1)(x-1) = (x-1)(x2)+(x-1)(2x)+(x-1)(1) = (x3-x2)+(2x2-2x)+(x-1) = x3+x2-x-1

So, in the end, the most simplified form of the original problem that you can get using basic concepts is:

ln(25x2/(x3+x2-x-1))

In theory, you could employ partial fraction decomposition to further break down the fraction inside of the natural logarithm function and possibly simplify it further, but this seems to be a problem from an Algebra II class, and partial fraction decomposition is not usually taught at that level in school, so I doubt you will need to break it down any further. If I am making false assumptions and you find out that you do indeed need to use partial fraction decomposition to break it down further, feel free to message me and I will edit this answer.