College Algebra Exercises Section 5.4

College Algebra Exercises Section 3.4

As you study these exercises, move your cursor over the light bulbs and the highlighted words.
Study with an active cursor

The definition that pops up when you move the cursor over a highlighted word is what should pop into your mind when you read, hear, or speak that word.

The material that pops up when you move the cursor over a light bulb is a suggested strategy for solving the problem. You should always formulate a similar strategy when you attempt to solve a problem.

When you move the cursor over an explanation of a step in a solution, the property that pops up is the mathematical justification for taking that action. You must always be able to provide such a justification for every step you take in mathematics.

17. Solve the equation ln(x) - ln(2) = 0 for x
Solution:
Add ln(2) to both sides of the equation to obtain
ln(x) = ln(2)
Apply the function exp to both sides of this equation to obtain
exp(ln(x)) = exp(ln(2))
Recall that exp and ln are inverses of each other to simplify this equation to
x = exp(ln(x)) = exp(ln(2))= 2
The solution is x = 2.

18. Solve the equation ln(x) - ln(5) = 0 for x
Solution:
Add ln(5) to both sides of the equation to obtain
ln(x) = ln(5)
Apply the function exp to both sides of this equation to obtain
exp(ln(x)) = exp(ln(5))
Recall that exp and ln are inverses of each other to simplify this equation to
x = exp(ln(x)) = exp(ln(5))= 5
The solution is x = 5.

19. Solve the equation e^x = 2 for x.
Solution: Recognize the left side of the equation as exp(x) so the equation to be solve is really exp(x) = 2.
Apply the ln function to both sides and use the fact that ln and exp are inverses of each other to obtain
x = ln(exp(x)) = ln(e^x)=ln(2)
Thus x = ln(2) which a calculator reveals to be 0.693
So the solution is 0.693.

20. Solve the equation e^x = 4 for x.
Solution: Recognize the left side of the equation as exp(x) so the equation to be solve is really exp(x) = 4.
Apply the ln function to both sides and use the fact that ln and exp are inverses of each other to obtain
x = ln(exp(x)) = ln(e^x)=ln(4)
Thus x = ln(2) which a calculator reveals to be 1.386
So the solution is 1.386.

21. Solve ln(x) = -1 for x
Solution:
Apply the exp function to both sides and use the fact that ln and exp are inverses of each other to obtain
x = exp(ln(x)) = exp(-1) = e^-1 = 0.368
Note: e^-1 may be computed directly using the e^x key on your calculator or by dividing 1 by an approximation of e.

22. Solve ln(x) = -7 for x
Solution:
Apply the exp function to both sides and use the fact that ln and exp are inverses of each other to obtain
x = exp(ln(x)) = exp(-7) = e^-7 = 0.0009 using the e^x key on your calculator

49. Solve e^3x = 12 for x
Solution:
Recognize the left side of the equation as exp(3x) so the equation to be solve is really exp(3x) = 12.
Apply the ln function to both sides and use the fact that ln and exp are inverses of each other to obtain
3x = ln(exp(3x) =ln(e^3x) = ln(12)
So 3x = ln(12) from which we obtain x =

50. Solve e^2x = 50 for x
Solution:
Recognize the left side of the equation as exp(2x) so the equation to be solve is really exp(2x) = 50.
Apply the ln function to both sides and use the fact that ln and exp are inverses of each other to obtain
2x = ln(exp(2x) =ln(e^2x) = ln(50)
So 2x = ln(50) from which we obtain

53. Solve 7 - 2e^x = 5 for x
Solution:
Begin by isolating the exponential expresion on one side ot the equation to get
2e^x = 2.
Divide both sies by 2 to obtain
e^x = 1
Clearly then x = 0.

57. Solve e^2x - 4e^x - 5 = 0
Solution:
Begin by recognizing this equation to be quadratic in form and let y = e^x to simplify the equation to
y² - 4y - 5 =0
Factor this quadratic to obtain
(y - 5)(y + 1) = 0
The Zero Factor Property permists us to infer that
y = 5 or y = -1
Substitute these possible values for y into the statement y = e^x and solve for x.
This gives us two equations
e^x = 5 and e^x = -1 to solve for x.
Using exp notation yields exp(x) = 5 and exp(x) = -1
Recall that the graph of exp(x) is above the x-axis for all values of x. That permits us to conclude the second equation (exp(x) = -1) has no solution.
To solve the equation exp(x) = 5 we apply the ln function to both sides of the equation and recall that exp and ln are inverses of each other.
This yields x = ln(exp(x)) = ln(5) = 1.609

87.
Solution:	Alternate Solution:

90. Solve

Solution:

92.Solve
Solution: