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Approximate the value of ln(1.1) using the tangent to f(x ... - Socratic 4 Oct 2017 · The value of ln(1.1) is approximately 0.1. We know that f(1) = ln(1) = 0. Furthermore, we know that f'(x) = 1/x. This means that f'(1) = 1. The equation of the tangent line is given by y …
How do you solve ln(lnx) = 1? - Socratic I found: x=e^e=15.154 You can use the definition of logarithm: log_ax=b->x=a^b and the fact that ln=log_e where e=2.71828...: we can write: ln(ln(x))=1 ln(x)=e^1 x=e^e=15.154
How do you simplify #ln(1/e) - Socratic 2 Mar 2018 · -1 Division rule of logarithms states that: ln(x/y) = ln(x) - ln(y) Here we can substitute: ln(1/e)=ln(1) - ln(e) 1) Anything to the power 0=1 2) ln(e)=1, as the base of natural logarithms is …
How do you find the Taylor series for ln(x) about the value x=1? 20 May 2015 · firstly we look at the formula for the Taylor series, which is: f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n which equals: f(a) + f'(a)(x-a) + (f''(a)(x-a)^2)/(2!) + (f ...
Find the general and principal value of log(-1+I)-log(-1-i)? - Socratic 17 Jun 2018 · We want to find: ln(-i) We know that the Polar form of a complex number is z=r(costheta+isintheta)=re^(itheta) => lnz = lnr+itheta Let z=-i. ln(-i) = lnr+itheta=ln| …
How do you solve lnx=-1? - Socratic 26 Jan 2017 · 1/e lnx=-1=>log_(e)x=-1 =>e^(-1)=x :.x=1/e. 139949 views around the world ...
How do you simplify ln 1/2? - Socratic 24 Mar 2016 · I found: -ln(2)=-0.69315 when the original question stated ln(1/2)...! I would use a property of the logs where you have: logx-logy=log(x/y) To write: ln(1/2)=ln(1)-ln(2)=0-ln(2)= …
How do you simplify #Ln(1/e^2)#? - Socratic 31 May 2018 · How do you simplify #Ln(1/e^2)#? Precalculus Properties of Logarithmic Functions Natural Logs. 1 Answer
What is the natural log of 1? - Socratic 5 Apr 2015 · The answer is 0. ln(1) is the same as asking e to what power is 1? Since anything to the 0 power is 1, ln(1) = 0
How do you evaluate #ln (1/e)#? - Socratic 10 Jun 2016 · It is -1. We apply the properties of the logarithm: ln(1/e)=ln(e^(-1)) the first property is that the exponent "exit" and multiply the log ln(e^-1)=-ln(e) the second property is that the …