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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 does 1 the l n mean?
The expression "1 the ln" appears to be a typographical error or a miscommunication. If you meant "1 ln," it could refer to the natural logarithm of 1, which is equal to 0, since the natural logarithm function (ln) answers the question: "To what power must e (approximately 2.718) be raised to yield 1?" The answer is 0, as e^0 = 1. If you meant something else, please provide clarification.
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 antiderivative of 1 over xlnx?
It is ln(ln(x))
