Q: What is 0.61 ln fraction?

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0.061 can be represented as the fraction 61/1000. As 61 is a prime number and is not a factor of 1000 then the fraction cannot be simplified and is thus already expressed in its lowest terms.

061 as a percentage = 6.1%0.061= 0.061 * 100%= 6.1%

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Take the natural logarithm (ln) of both sides of the equation to cancel the exponent (e). For example, ify=Aexlog transform both sides and apply the rules of logarithms:ln(y)=ln(Aex)ln(y)=ln(A)+ln(ex)ln(y)=ln(A)+xrearrange in terms of x:x=ln(y)-ln(A), or more simplyx=ln(y/A)

2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)

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0.061 can be represented as the fraction 61/1000. As 61 is a prime number and is not a factor of 1000 then the fraction cannot be simplified and is thus already expressed in its lowest terms.

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.

Ln 4 + 3Ln x = 5Ln 2 Ln 4 + Ln x3= Ln 25 = Ln 32 Ln x3= Ln 32 - Ln 4 = Ln (32/4) = Ln 8= Ln 2

061 as a percentage = 6.1%0.061= 0.061 * 100%= 6.1%

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ln(ln)

Take the natural logarithm (ln) of both sides of the equation to cancel the exponent (e). For example, ify=Aexlog transform both sides and apply the rules of logarithms:ln(y)=ln(Aex)ln(y)=ln(A)+ln(ex)ln(y)=ln(A)+xrearrange in terms of x:x=ln(y)-ln(A), or more simplyx=ln(y/A)

Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x

2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)

You can also write this as ln(6 times 4)

3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)

It depends. If you mean (ln e)7, then the answer is 1, since (ln e) = 1. If you mean ln(e7), then the answer is 7, since ln(e7) = 7 (ln e) and (ln e) = 1.