Here is another way to do the integral. Observe that the area is equivalent to the area below #y=e^x# (the inverse function) from #x=-oo# to #0# but opposite in sign. #-int_ (-oo)^0e^x\ dx# …

Rule 1: sqrt {"nonnegative"} 7-ln (x-1) ge0 by adding ln (x-1), => 7 ge ln (x-1) by raising e to both sides, => e^7 ge e^ {ln (x-1)}=x-1 by adding 1, => e^7+1 ge x Rule2: ln ("positive") x-1 > 0 by …

Explanation: This integral may seem tricky at first, but it comes apart quite quickly once you realize a certain u-substitution. If we let #u=ln (ln (x))#, we get by the chain rule that the …

Please see the explanation. Let y = sqrt (x)^x Because the square root is the same as the 1/2 power, we can write the above equation as: y = (x^ (1/2))^x A property of exponents allows us …

#e^ln (sin (2x))= e^0; 0 < x < pi/2# The exponential function and the natural logarithm are inverses, therefore, they cancel and any number to the 0 power is 1:

You can do that by writing the Arrhenius equation in non-exponential form ln (k) = ln [A * "exp" (-E_a/ (RT))] This is equivalent to ln (k) = ln (A) + ln ["exp" (-E_a/ (RT))] ln (k) = ln (A) - E_a/ …

Let us find y'. By repeatedly applying Chain Rule, y'=2 [ln (1+e^x)]^1cdot [ln (1+e^x)]' =2ln (1+e^x)cdot {1}/ {1+e^x}cdot (1+e^x)' =2ln (1+e^x)cdot {1}/ {1+e^x}cdot ...

May 4, 2017 · To have these counts the percentage has to be different. Approximately 5.7% ~~3" years "69 1/4" days" color (blue) ("Preamble about the model to use") You have what I ...

See below. If you are looking for the area bounded by y=e^ (2x) and y=5, it will be between the curve, the line y =5 and the y axis. This will be in the interval x in [ 0 ,1 ] You first need to …

The integrated rate law for a first-order reaction A → products looks like this ln([A]t [A]0) = − k ⋅ t −−−−−−−−−−−−−−−−−− Here [A]t is the concentration of the reactant after a period of time t …