"Ln" in that equation is the "natural logarithm" of a number.
The "common logarithm" ... log(x) ... is the logarithm of 'x' to the base of 10.
The "natural logarithm" ... ln(x) ... is the logarithm of 'x' to the base of 'e'.
'e' is an irrational number, known, coincidentally, as the "base of natural logarithms".
It comes up in all kinds of places in math, physics, electricity, and engineering, especially in
situations where the speed of something depends on how far it still has to go to its destination.
'e' is roughly 2.7 1828 1828 45 90 45 ... (rounded)
The entropy equation, S k ln W, is used in thermodynamics to quantify the amount of disorder or randomness in a system. Here, S represents entropy, k is the Boltzmann constant, and W is the number of possible microstates of a system. By calculating entropy using this equation, scientists can measure the level of disorder in a system and understand how it changes over time.
Actual vapor pressure can be calculated using the Antoine equation, which is a function of temperature and constants specific to the substance of interest. The equation is: ln(P) = A - (B / (T + C)), where P is the actual vapor pressure, T is the temperature in Kelvin, and A, B, and C are substance-specific constants.
Water vapor pressure can be calculated using the Clausius-Clapeyron equation, which relates vapor pressure to temperature. The equation is: ln(P2/P1) (Hvap/R)(1/T1 - 1/T2), where P1 and P2 are the vapor pressures at temperatures T1 and T2, Hvap is the heat of vaporization, and R is the gas constant.
In statistical mechanics, the Helmholtz free energy is related to the partition function through the equation F -kT ln(Z), where F is the Helmholtz free energy, k is the Boltzmann constant, T is the temperature, and Z is the partition function. This equation describes how the Helmholtz free energy is connected to the microscopic states of a system as described by the partition function.
The electric potential due to an infinite line charge decreases as you move away from the charge. The formula to calculate the electric potential at a distance r from the line charge is V / (2) ln(r), where is the charge density of the line charge, is the permittivity of free space, and ln(r) is the natural logarithm of the distance r.
ln 60 = a
If the equation was ln(x) = 2.35 then x = 10.4856, approx.
The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)
To find ln 2.33, you need a calculator. It is the solution of the equation e^x = 2.33. ln 2.33 = 0.84586 (using a calculator)
In the equation ln(x) = 5, the solution is x = (about) 148.4. To solve, simply raise e to the power of both sides and reduce... ln(x) = 5 eln(x) = e5 x = 148.4
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)
The correct formula for exponential interpolation is: y =ya*(yb/ya)^[(x-xa)/(xb-xa)], xa<x<xb and also, x=xa*[ln(yb)-ln(y)]/[ln(yb)-ln(ya)]+xb*[ln(y)-ln(ya)]/[ln(yb)-ln(ya)], ya<y<yb
yes
To be logarithmic the equation would be, ( otherwise just linear ) 3x + 7 = 20 subtract 7 from each side 3x = 13 now, you know the answer is between x = 2 and x = 3, so I use natural logs both sides ln(3x) = ln(13) as this is a logarithmic operation you can bring down the x in front of the ln sign on the left x ln(3) = ln(13) divideboth sides by ln(3) x = ln(13)/ln(3) ( not ln(13/3)!!!!! ) x = 2.334717519 -------------------------check in original equation 3(2.334717519) + 7 = 20 13 + 7 = 20 20 = 20 --------------checks
For the function: y = x^x^x (the superscript notation on this text editor does not work with double superscripts) To solve for the derivative y', implicit differentiation is needed. First, the equation must be manipulated so there are no x's raised to x's on the right side of the equation. So, both sides of the equation must be input into a natural logarithm, wherein we can use the properties of logarithms to remove the superscripted powers of the right side: ln(y) = ln(x^x^x) ln(y) = xxln(x) ln(y)/ln(x) = xx ln(ln(y)/ln(x)) = xln(x) eln(ln(y)/ln(x)) = exln(x) ln(y)/ln(x) = exln(x) ln(y) = ln(x)exln(x) Now there are no functions raised to functions (x's raised to x's). Deriving this equation yields: (1/y)(y') = ln(x)exln(x)(x(1/x) + ln(x)) + exln(x)(1/x) = ln(x)exln(x)(1 + ln(x)) + exln(x)(1/x) = exln(x)(ln(x)(1+ln(x)) + (1/x)) Solving for y' yields: y' = y[exln(x)(ln2(x) + ln(x) + (1/x))] or y = xx^x ln(y) = ln(x)x^x ln(y) = xxln(x) ln(y) = exlnxln(x) y'/y = exlnx[ln(x) + 1)ln(x) + exlnx(1/x) y' = y[exlnx(ln2(x) + ln(x) + 1/x)] y' = xx^x[exlnx(ln2(x) + ln(x) + 1/x)]
A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)
the natural log, ln, is the inverse of the exponential. so you can take the natural log of both sides of the equation and you get... ln(e^(x))=ln(.4634) ln(e^(x))=x because ln and e are inverses so we are left with x = ln(.4634) x = -0.769165