Review the following concepts if needed:
 Logarithm Basics Help
 Derivative of Logarithm Function Help
 Exponential Basics Help
 Derivative of Exponential Help
 Exponentials with Arbitrary Bases Help
 Logarithms with Arbitrary Bases Help
 Derivative and Integral of Logarithm Help
 Logarithm and Exponential Graphing Help
 Logarithmics Differentiation Help
 Exponential Growth and Decay Help
 Radioactive Decay Help
 Compound Interest Help
 Inverse Trigonometric Functions Help
 Other Inverse Trignometric Functions Help
Transcendental Functions Practice Test

Simplify these logarithmic expressions.

Solve each of these equations for x .

Calculate each of these derivatives.

Calculate each of these integrals.
(a) ∫ e ^{−x} x ^{3} dx [ Hint : Guess p ( x ) · e ^{−x} , p a polynomial.]

Use the technique of logarithmic differentiation to calculate the derivative of each of the following functions.

There are 5 grams of a certain radioactive substance present at noon on January 10 and 3 grams present at noon on February 10. How much will be present at noon on March 10?

A petri dish has 10,000 bacteria present at 10:00 a.m. and 15,000 present at 1:00 p.m. How many bacteria will there be at 2:00 p.m.?

A sum of $1000 is deposited on January 1, 2005 at 6% annual interest, compounded continuously. All interest is reinvested. How much money will be in the account on January 1, 2009?

Calculate these derivatives.

Calculate each of these integrals.
Answers


We do (a) and (b).



We do (a) and (b).
(a) Let . Then
ln A = 3 ln x + ln( x ^{2} + 1) − ln( x ^{3} − x )
hence
Multiplying through by A gives
(b) Let . Then
ln A = ln sin x + ln( x ^{3} + x ) − ln x ^{2} − ln( x + 1)
hence
Multiplying through by A gives

Let R ( t ) denote the amount of substance present at time t . Let noon on January 10 correspond to t = 0 and noon on February 10 correspond to t = 1. Then R (0) = 5 and R (1) = 3. We know that
R ( t ) = P · e ^{Kt} .
Since
5 = R (0) = P · e ^{K · 0} ,
we see that P = 5. Since
3 = R (1) = 5 · e ^{K · 1} ,
we find that K = ln 3/5. Thus
Taking March 10 to be about t = 2, we find that the amount of radioactive material present on March 10 is

Let the amount of bacteria present at time t be
B ( t ) = P · e ^{Kt} .
Let t = 0 be 10:00 a.m. We know that B (0) = 10000 and B (3) = 15000. Thus
10000 = B (0) = P · e ^{K · 0}
so P = 10000. Also
15000 = B (3) = 10000 · e ^{K · 3}
hence
As a result,
B ( t ) = 10000 · e ^{t · [1/3] ln(3/2)}
or
We find that, at 2:00 p.m., the number of bacteria is

If M ( t ) is the amount of money in the account at time t then we know that
M ( t ) = 1000 · e ^{6 t /100} .
Here t = 0 corresponds to January 1, 2005. Then, on January 1, 2009, the amount of money present is
M (4) = 1000 · e ^{6 · 4/100} ≈ 1271.25.


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