2017年10月1日 · It does not matter what base we use providing the same base is used for all logarithms, here we are using bease e. Let us define A,B.C as follows=: A = lna ⇔ a = eA, B = lnb ⇔ b = eB C = ln(a b) ⇔ a b = eC From the last definition we have: a b = eC ⇒ eC = eA eB And using the law of indices: eC = (eA)(e−B) = eA−B

2020年6月2日 · My calculus book states the following theorem of the properties of natural logarithms: If a, b > 0 , then ln(ab)= lna + lnb The author goes on to prove this theorem as follows I do not understan...

2022年11月16日 · Im reading the book: What Is Mathematics An Elementary Approach to IDEAS AND METHODS. There is a description before the proof of ln a + ln b = ln(ab) ln a + ln b = ln (a b) Intuitively, this formula could be obtained by looking at …

2017年6月25日 · I know that formula, but I don't understand it. $\\log_a(b)=\\frac{\\ln(b)}{\\ln(a)}$ Thanks.

But how can I continue from here? The original question in my textbook asked if the equality ln(a + b) = ln(a) + ln(b) ln (a + b) = ln (a) + ln (b) was true. My initial answer was that it was not true, but I missed this special case, which I think was very hard to …

2023年10月5日 · I have seen proofs for lnab = lna + lnb ln a b = ln a + ln b that use the integral definition of lnx ln x but is there a proof that uses the limit definition

2017年10月29日 · lna = lnb Clearly {a and b}> 0 for this equation to be valid. Raise e to the power of each side of the equation. elna = elnb Remember that elnx = x Thus: elna = a and elnb = b Hence, a = b

2016年8月23日 · 20lna = lna20 by the rule bloga = logab. Similarly, 4lnb = lnb4 by the same rule. So now we have 20lna −4lnb = lna20 −lnb4. Now, using the rule loga −logb = log(a b), we can rewrite this as ln(a20 b4). If you like, you can write this as ln((a5 b)4), which can be rewritten as 4ln(a5 b). You can plug numbers in to these rules to make sure they work. For example, log232− log28 = 5 − 3 ...

2016年11月6日 · Prove that aln(b) =bln(a) a ln (b) = b ln (a) for any values of a a and b b that are positive and different from 1. My book says the solution is:

2021年1月25日 · I think, since this is equivalent to a + b ≤ ab ⇔ 1 ≤ b − b a a + b ≤ a b ⇔ 1 ≤ b − b a it holds for 1 <b <a 1 <b <a only and, moroever, it is a strict inequality for these values.