0
ln 1 = 0
e0=1
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What is the value of loge if base is not equal to e?
You can calculate log to any base by using: logb(x) = ln(x) / ln(b) [ln is natural log], so if you have logb(e) = ln(e) / ln(b) = 1 / ln(b)
What is the derivative of y equals xlnx?
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
Integral of ln2x?
I will denote an integral as \int (LaTeX). We can let u = 2x and du = 2dx, and substitute \int ln (2x) dx = (1/2) \int ln u du. Either using integration by parts or by memorization, this is equal to (1/2) u ln u - u + C = (1/2)(2x ln (2x) - 2x) + C, where C is an arbitrary constant.
How can you prove that yis equal to e raise to the power xlny?
Euler's constant, e, has some basic rules when used in conjunction with logs. e raised to x?æln(y),?æby rule is equal to (e raised to ln(y) raised to x). e raised to ln (y) is equal to just y. Thus it becomes equal to y when x = 1 or 0.
Does the series 1 divided by ln x converge?
Compare a series to a known series. So take the harmonic series {1/1 + 1/2 + 1/3 + ... + 1/n}, which diverges.For each number n [n>1], LN(n) < n, so 1/(LN(n)) > 1/n. So since each term in 1/LN(n) is greater than each term in the divergent series {1/n}, then the series 1/LN(n) diverges.
What is the derivative of ln 1?
It is equal to 0
What is the value of loge if base is not equal to e?
You can calculate log to any base by using: logb(x) = ln(x) / ln(b) [ln is natural log], so if you have logb(e) = ln(e) / ln(b) = 1 / ln(b)
Why does lne equal 1?
e = 2.71828183 (approximately)The definition of ln is this: ln x = y when e ^ y = x. It's an inverse property... So ln x means "find out what value y would need to have so that e ^ y equals x" Since e ^ 1 = e, ln e has to equal 1. because in line equation to signify that the task/job is done. This is why it is equal to 1.cause you add them and it just does
What is the derivative of y equals xlnx?
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
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.
Integral of ln2x?
I will denote an integral as \int (LaTeX). We can let u = 2x and du = 2dx, and substitute \int ln (2x) dx = (1/2) \int ln u du. Either using integration by parts or by memorization, this is equal to (1/2) u ln u - u + C = (1/2)(2x ln (2x) - 2x) + C, where C is an arbitrary constant.
What is the answer to 2 minus Ln times 3 minus x equal 0?
so, if 2 minus Ln times 3 minus x equals 0, then 2 minus Ln times 3 equals x, therefore 2 minus Ln equals x divided by three, so Ln + X/3 = 2 therefore, (Ln + [X/3]) = 1
Do any positive real numbers alpha and beta exist such that lnalpha times beta equals lnalpha plus lnbeta and if so what are they?
I assume the question is NOT about ln(a*b) = ln(a) + ln(b) because that is true for all positive real a and b. Instead, you want a solution to ln(a) * b = ln(a) + ln(b) or, ln(a) * (b-1) = ln(b) ln(a) = ln(b)/(b-1) ln(a) = ln[b1/(b-1)] Exponentiating, a = b1/(b-1) For any real number b > 1, a given by the above equation will meet your requirements.
Why does ln x plus 2 equal 2 plus ln x?
Because of the commutative property of addition.
What is the antiderivative of 1 over xlnx?
It is ln(ln(x))
How many prime between 1 and 100?
25. Note: The number of primes up to a certain integer "x" is roughly equal to x / ln(x), where ln(x) is the natural (base-e) logarithm.
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)]
