|
DrDelMath College Algebra
Examples |
As
you study these exercises, move your cursor over owls, light bulbs, highlighted
words, and equations.
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. |
Problem: 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.
Problem: 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.
Problem: Solve the equation ex = 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(ex)=ln(2)
Thus x = ln(2) which a calculator reveals to be 0.693
So the solution is 0.693.
Problem: Solve the equation ex = 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(ex)=ln(4)
Thus x = ln(2) which a calculator reveals to be 1.386
So the solution is 1.386.
Problem: 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 ex key on your calculator or by dividing 1 by an approximation of e.
Problem: 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 ex key on your calculator
Problem: Solve e3x = 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(e3x) = ln(12)
So 3x = ln(12) from which we obtain 
Problem: Solve e2x = 50 for x
Solution:
Recognize the left side of the equation as exp(2x) so the equation to be solved 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(e2x) = ln(50)
So 2x = ln(50) from which we obtain 
Problem: Solve 7 - 2ex = 5 for x
Solution:
Begin by isolating the exponential expresion on one side ot the equation to get
2ex = 2.
Divide both sies by 2 to obtain
ex = 1
Clearly then x = 0.
Problem:Solve e2x - 4ex - 5 = 0
Solution:
Begin by recognizing this equation to be quadratic in form and let y = ex 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 = ex and solve for x.
This gives us two equations
ex = 5 and ex = -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
 |
|
Solution:  |
Alternate Solution: |
Solve 
Solution:
 |
Solve |
