Q: Evaluate the function e squared by ln 3?

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2b=6squared, and 6x6=36

Y = x squared -4x plus 3 is an equation of a function. It is neither called a domain nor a range.

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

e^(3lnx)=e^[ln(x^3)]=x^3

3 squared=9 3 squared=3*3 3*3=9

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3 squared = 9 9 - 2 x 1 = 9 - 2 = 7

2b=6squared, and 6x6=36

-1/3,1/3and 5/3

Yes.

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

It is 1.2164

An exponential function can be is of the form f(x) = a*(b^x). Some examples are f1(x) = 3*(10^x), or f2(x) = e^(-2*x). Note that the latter still fits the format, with b = e^(-2). The inverse is the logarithmic function. So for y = f1(x) = 3*(10^x), reverse the x & y, and solve for y:x = 3*(10^y)log(x) = log(3*(10^y)) = log(3) + log(10^y) = log(3) + y*log(10) = y*1 + log(3)y = log(x) - log(3) = log(x/3)The second function: y = e^(-2*x), the inverse is: x = e^(-2*y).ln(x) = ln(e^(-2*y)) = -2*y*ln(e) = -2*y*1y = -ln(x)/2 = ln(x^(-1/2))See related link for an example graph.

dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).

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

Y = x squared -4x plus 3 is an equation of a function. It is neither called a domain nor a range.

X = 1.31356+0.612045*iSteps to solve, take the natural log of both sides:ln(X^(3-5i)) = ln(23-14i).(3-5i)*ln(X) = ln(23-14i). Convert 23-14i to exponential form: A*e^(iÃŽËœ) {A = 26.926 and ÃŽËœ = -0.54679 radians}(3-5i)*ln(X) = ln(A*e^(iÃŽËœ))= ln(A) + iÃŽËœ = ln(26.926) - 0.54679i.divide by (3-5i): ln(X) = (ln(A) + iÃŽËœ) / (3-5i) = (3.2931 - 0.54679i)/(3-5i)So we have ln(X) = 0.370978 + 0.436033i, then:e^(ln(X)) = e^(0.370978 + 0.436033i) --> X = 1.31356+0.612045*i

e^(3lnx)=e^[ln(x^3)]=x^3