
How do you simplify #ln(1/e) - Socratic
Mar 2, 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 always e Here, we can simplify: ln(1)=0 ln(e)=1 Thus: ln(1)-ln(e)=0-1 =-1 Thus, we have our answer
How do you evaluate #ln (1/e)#? - Socratic
Jun 10, 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 logarithm of the base is 1. The base of the natural logarithm is e then …
How do you simplify #Ln(1-e^-x)#? - Socratic
Feb 25, 2017 · From here, use #ln(a/b)=ln(a)-ln(b)# and #ln(e^x)=x#: #=ln(e^x-1)-ln(e^x)=color(blue)(ln(e^x-1)-x# I don't know if this is a simplification per se, but it's definitely a valid way to rewrite the function.
How do you simplify #ln (1/e^3)#? - Socratic
Dec 3, 2015 · ln(1/e^(3))=ln(e^(-3))=-3. Since ln(x) and e^{x} are inverse functions, ln(e^{x})=x for all values of x. Since 1/e^{3}=e^{-3} by definition of negative exponents, it ...
How do you simplify #Ln(-e) - ln(-1/e)#? - Socratic
Mar 23, 2016 · This expression is not valid because #ln# is not defined for negative numbers but if it was:. #ln(e)-ln(1/e)# then you could use a formula: #lna-lnb=ln(a/b)# which would lead to:
How do you simplify #Ln(1/e^2)#? - Socratic
May 31, 2018 · How do you simplify #Ln(1/e^2)#? Precalculus Properties of Logarithmic Functions Natural Logs. 1 Answer
How do you simplify e^-lnx? + Example - Socratic
Feb 6, 2016 · #color(brown)("Total rewrite as changed my mind about pressentation.")# #color(blue)("Preamble:")# Consider the generic case of #" "log_10(a)=b#
How do you simplify e^lnx? - Socratic
Mar 22, 2016 · e^lnx=x let y=e^lnx ln y=lne^lnx->Take ln of both sides lny = lnx * ln e -> use the property log_b x^n = nlog_b x lny=lnx(1)-> ln_e e = 1-> from the property log_b b = 1 lny = ln x Therefore y=x
What is the natural log of 1? - Socratic
Apr 5, 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 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