Monday, January 23, 2012, Math 152-06
Section 6.4. General Exponential and Logarithm Functions
Section 6.5. Exponential Growth and Decay

<<PlotLegends` (*Needed to put legends inside a plot*)

Next is plotted the general exponential function y=(a^x) versus its inverse y=(Subscript[log, a])x with the line y=x and a square window with aspect ratio 1. Press the play button to see the graph change with a, but you may need to slow down the animation using the multi-down arrow button.

Manipulate[Plot[{x,a^x,Log[a,x]},{x,-2,6},PlotStyle->{Dashed,Thick,Thick},PlotRange->{-2,6},AspectRatio->1,PlotLegend->{"y=x","(a^x)","(Subscript[log, a])(x)"},LegendPosition->{.8,.2}],{{a,E},.1,3,.1}]

Interestingly, for some values of a, the graphs intersect. For a=1.4, we can use the FindRoot function (think Newton’s Method) to estimate the x values at which the graphs intersect. Recall that Newton’s Method starts at an initial x value and iterates based on the tangent line at the current x value to refine its estimate of the root.

(* Form: FindRoot[LHS - RHS,{x,initial x}] *)
FindRoot[1.4^x-Log[1.4,x],{x,2}]
FindRoot[1.4^x-Log[1.4,x],{x,6}]

Exercise

Find the (exact or approximate to 5 decimal places) value of a, a>1.4, for which the graph of y=a^x and y=Log[a,x] intersect in exactly one point. Try to figure out how to use FindRoot (Newton’s method) for this.

Example.

What Mathematica returns for the general logarithm antiderivative: ln(x) times a constant. Mathematica leaves out the absolute value bars in ln(x) even though it is a valid formula to keep them.

Integrate[(1/Log[10])*(1/x),x]

Example graph of function and inverse

We algebraically solved for the inverse of the function y=(10^x)/((10^x)+1) and plotted the function and the inverse.

Solve[y==(10^x)/((10^x)+1),x] (*This is equivalent to what we solved on the board.*)

Plot[{x,10^x/(10^x+1),Log[10,x/(1-x)]},{x,-3,3},PlotRange->{-3,3},AspectRatio->1,PlotStyle->{Dashed,Thick,Thick},PlotLegend->{"y=x","f(x)","\!\(\*Cell[TextData[Cell[BoxData[
FormBox[
RowBox[{
SuperscriptBox["f", 
RowBox[{"-", "1"}]], "(", "x", ")"}], TraditionalForm]],
FormatType->"TraditionalForm"]]]\)"}]

Exponential Growth and Decay

The examples below are applications of the differential equation
                             dy/dt = k y
and its general solution
			y(t)=y(0) (e^(k t)).
We discussed that y[0] is the initial amount/population. Also, by rewriting the differential equation, we get
			(dy/dt)(1/y)=k
so that k is interpreted as the rate of change relative to the total, or the percent rate of change.

Example. Population growth in the US.

See your notes. Data was: year 1900, population 76 (million). Year 2000, population 281 (million). We obtained the general expression for P(t):

P[t_]:=76 (E^(k t))
P[t]

Now we obtain k by solving P[100]=281 for k:

Solve[P[100]==281,k]
%//N (*Numerical approximation*)
Print["If it were constant, the percent population growth rate from 1900-2000 would be: " ,100*%[[1]][[1]][[2]],"%"]

m[t_]:=C (E^(k t))
Solve[m[t]==1/2 C,t]

Radioactive Decay

m(t)  mass of radioactive material at time t
(Subscript[m, 0]) initial mass of material
decay rate (as a fraction of material) is ((-1)/m)(dm/dt).
General solution: m(t)=m(0)(e^(k t)), where k is a negative value.
Defn. Half life is the amount of time for exactly half of the substance to decay.
Half life formula (solve m(t)=m(0)/2  for t.)
		t=((ln(1/2))/k)

Example: Fukushima disaster

Cesium-137: Half life of 30.17 years.
1.1x10^16 Bequerels of Cesium-137 released.  1 Bequerel is whatever amount of substance needed so that 1 decay occurs per second.
(a) formula for m(t)
(b) How much left after 200 years
(c) When will 1% of original amount be left?
(d) (on your own) How bad is Cesium-137 pollution? What is the average Bequerels per square mile of Cesium-137 radiation in background radiation before Fukushima? Estimate how much worse (as a multiplier above average) the radiation has been locally around the Fukushima plant in the time after the disaster.

m[t_]:=1.1*10^16 E^(k t)
m[t]

From half life data we get k by solving:

m[30.17]==m[0](1/2)
Solve[m[30.17]==m[0](1/2),k]

(a) Now with both k and m(0) we can plot the level of cesium-137 for all t.

m[t_]:=1.1*10^16 E^(-0.022974715961549395` t)

Plot[m[t],{t,0,200},PlotRange->{0,1.1*10^16},AxesLabel->{"time in years","undecayed Cesium-137"}]

(b) How much after 200 years?

m[t]
m[200]

(c) When will 1% be left?

.01*m[0]==m[t]
Solve[.01*m[0]==m[t],t]