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# Transcendental Functions Practice Test

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By — McGraw-Hill Professional
Updated on Sep 1, 2011

Review the following concepts if needed:

## Transcendental Functions Practice Test

1. Simplify these logarithmic expressions.

2. Solve each of these equations for x .

3. Calculate each of these derivatives.

4. Calculate each of these integrals.

(a) ∫ e −x x 3 dx [ Hint : Guess p ( x ) · e −x , p a polynomial.]

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

6. 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?

7. 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.?

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

9. Calculate these derivatives.

10. Calculate each of these integrals.

1.

2. We do (a) and (b).

3.

4.

5. We do (a) and (b).

(a) Let . Then

ln A = 3 ln x + ln( x 2 + 1) − ln( x 3x )

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

6. 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

7. 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

8. 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.

9.

10.

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