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Simplify In e to the 7th power?
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.
What does Ln mean?
in math, ln means natural log, or loge and e means 2.718281828
What is the derivative of lnx2?
Do you mean ln(x-2), or ln(x)-2? If it is ln(x-2): 1/(x-2) If it is ln(x)-2: 1/x
What does LN mean in light bulbs?
long neck
What does LN from away we go mean?
Lactating nurse
How do you solve for the x power of 8 is 20?
if x8 = 20 then x = - eighth root of 20 or x = eighth root of 20 eighth root of 20 = 1.4542154.... Maybe you mean 8x = 20 If so Ln (8x) = Ln (20) => x Ln(8)= Ln(20) => x = Ln(20)/Ln(8) = 1.4406426...
How would you solve ln 4 plus 3 ln x equals 5 ln 2?
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
What does ln2 equal?
Assuming you mean 'logarithm to the base 'e' ( natural logarithm. On the calculator its symbol is 'ln'. Hence ;ln 2 = 0.69314718....
What is the derivative of Ln 2 plus x?
The order of operations is not quite clear here.If you mean (ln 2) + x, the derivate is 0 + 1 = 1.If you mean ln(2+x), by the chain rule, you get (1/x) times (0+1) = 1/x.
How do you work out Ln 24 - ln x equals ln 6?
18
What is the dervative of x pwr x pwr x?
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)]
Why is the symbol for natural log ln?
ln(ln)
