If y is a rational number and x a positive number, ln x y = y ln x. This is proofed by letting y to be a rational number m/n where n a positive number. Then ln x y = ln x m/n = ln(√n x) m = m ln √n x = m n ln x = y ln x. A logarithmic function written as y = l x, has an implied base of 10. The natural logarithmic function y ln x is equal to a logarithm function with a base e which is thought of as e to the y power equals x and e is approximately 2.718. The natural logarithms system is different from the system of common logarithms which have 10 as the base and are used in most practical work. The base e denotes the y ln x function. logex is equal to ln x and y ln x is written as e y which means the logarithm function y ln x inverse is the exponential function y = ex. The inverse relationships are ln ex = x and eln x = x. And the logarithm of the base itself is usually 1: ln e = 1.The function y ln x is continuous and defined for every positive values of x. This function will obey the usual rules that govern logarithms. These rules that govern how logarithms work are:

- logbxy = logbx + logby: This is the y ln x product rule which states that logarithms of a products are equal to the sum of the logarithms of each factor. For example ln(3)( 7) = ln(3) + ln(7).
- logb x/y= logbx − logby: This is the quotient rule which states that logarithms of quotients is equal to the logarithm of the numerator minus the logarithm of the denominator. For example ln(3 / 7) = ln(3) - ln(7).
- logb xn = n logbx or ln(x^y) = y ln x : This is the power rule which states that the logarithm of x with a rational exponent is equal to the exponent multiplied by the logarithm. For example ln(2^8) = 8 x ln(2).

And like all algebra rules, y ln x will obey the rule of symmetry. For example, n ln a = ln an. The derivative of y ln x applies the definition of a derivative to prove: d/dx lnx=1/x and when proofing it is necessary to define the base of the system of natural logarithms which is the number e, as the limit: , y = 1 if the base a is equal to the value of x.

The y ln x function is a logarithm and all logarithms graphs look alike. To plot the y ln x graph, you begin by building a table of values and then plot the points until when you have a perfect idea of how you will connect the dots to form a smooth curve. X is an argument of y ln x and arguments of y ln x function no matter their base, they must be positive. This means that domain will contain positive numbers only. This implies that when building a table of values for plotting the y ln x graph you only select the positive numbers of x.

When you have a calculator which has the ln button, you will not need to be selective of what x positive values to choose when building the table of values for the y ln x graph. But if you do not have the calculator, you will choose the x values whose ln can be determined by simple hand calculations. Similarly if your calculator does not have the ln function, choose x values that are known powers of ln base. For example e^0=1 which means ln 1 is equal to zero thus making the point (1, 0) a point of the y ln x graph. In the same way e^2 is equal to 1 and so the point (e, 1) is also a point on the y ln x graph and so forth. For the number e, you will use 2.8 for approximation. Note that if you use a calculator program which has windows, you can easily find y ln x with it. And if the button of ln is not available upon starting the program, you can click on the view menu and select scientific. This will display more buttons and you can find y ln x.

You can also find the equation of a tangent of a line with the y ln x function. To get the equation of a tangent line to at y ln x at the point x=1, you will simply use the basic slope y intercept equation. Let f(x) = lnx. Then f(1) = 0, f'(x) = 1/x and f'(1) = 1. This is to say let the point be (a, f(a)) instead of (x1, f(x1)) and m = f'(a) and since x = 1, let x1 = a = 1, you will use y - y1 = m(x - x1) as y = f(a) + f'(a)(x - a). This will make the equation: y = f(1) + 1(x - 1) ==> or y = 0 + x – 1. Estimation will use the equation y = x – 1 implying that ln(1.1) ≈ 1.1 - 1 = 0.1, and ln(2) ≈2 - 1 = 1.

If you are given ln(y)=ln(x)+ln(c) and you need to solve for y, use the y ln x product rule to solve it. It will become logb(x)+logb(y)=logb(xy). ln(y)=ln(xc). Place all terms that have a logarithm to the left side of the equation. ln(y)−ln(xc)=0. Then use the y ln x quotient rule, logb(x)−logb(y)=logb(xy). ln(yxc)=0. You will then rewrite ln(yxc)=0 in exponential form using the definition of a y ln x function. If b and x are positive real numbers and b is not equal to 1, then logb(x)=y is equivalent to by=x. (e)0=yxc. Now remove the brackets in e. e0=yxc anything raised to 0 is 1. 1=yxc. And since y is on the right side of the equation, switch sides in order to have it on the left side of the equation. yxc=1. Then multiply both sides of the equal sign: xc. y=1⋅(xc) and simplify the right side. Multiply 1 by xc to get 1(xc). y=1(xc) multiply x by 1 to get x. y=xc write down x and c. y=cx. The equation will be y=ln(x2 -3x+1).

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