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Write the expression as a single logarithm of a single quantity 2lnx-2lnx plus 1 - lnx-1 plus 2ln5?
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.
Simplify log 1 over 2 times 8?
[(log1)/2](8)=(0/2)(8)=0
Simplify 2 over 6?
8
What is the derivative of ln 1-2x?
3
What is the answer to 2 minus Ln times 3 minus x equal 0?
so, if 2 minus Ln times 3 minus x equals 0, then 2 minus Ln times 3 equals x, therefore 2 minus Ln equals x divided by three, so Ln + X/3 = 2 therefore, (Ln + [X/3]) = 1
