4.04  Solving Exponential Equations

with Logarithms

College Algebra: One Step at a Time. 

Page 523:   #3, 16, 17, 19, 23, 27, Extra Prob, 29, 33, 34, 35, 36, Extra Prob.

Dr. Robert J. Rapalje

Seminole State College of Florida

Sanford, FL  32773

To see Section 4.04, with detailed explanations, examples, exercises, and answers, click here!

Solving equations using logarithms is not nearly as hard as you think it is, especially if it is explained in living color!  Consider these exercises from pages 520-523.  Notice how the colors make the exercises so much easier to follow.  Can you imagine what this would look like in black and white?  Most of our television is in color--why not math?  Have your graphing calculator ready, especially if you have a TI 84 or a TI 83+!

P. 521 # 3.  Three Solutions for 

First solution:  Laws of exponents--a "contrived" problem. 

Begin by noticing that this is a very “special” problem in which both sides of the equation can be written as a power of the same number 3.  In mathematics we call this a “contrived problem,” one that we made up especially because of this special property, and therefore the method is of little or no use to real life problems, where things usually don’t come out even. 

Notice that  and , and substitute these into the original problem:

Remember that when you raise a power to a power, you multiply exponents:

Notice that the base number on both sides of the equation is the same.  I call it the

                        “This Equals That Theorem”:

                   If , then

Solve the equation by subtracting 4x and 6 from each side:

Divide both sides by -1:

Second solution:  Using Logarithms. 

How would you solve an exponential equation if it does NOT have the same base number on both sides of the equation?

Take the ln of both sides of the equation:

Using the second law of logarithms:

Using the distributive property:

Get all x terms on the left side by subtracting    and from both sides of the equation :

Get the x variable in one place by factoring out the common factor of  x:

Solve for x by dividing each side of the equation by  

In the final step, parentheses are included that would be suitable for calculating the answer with a TI 83 or TI 84.  To calculate with a TI 85 or TI 86, use parentheses as indicated in the previous step.  Notice that for ALL calculators, you must have parentheses around the numerator and the denominator.  Notice also that at the beginning of the numerator, it is a NEGATIVE of the ln 9, but the other two signs in the problem are MINUS (not negative!) signs.  Also, for the TI 83/84 calculator, be reminded that there must be a DOUBLE closed parentheses at the end of the numerator!

Calculate with your calculator:

                                                x = 8

Third solution: Using a Graphing Calculator—TI84 (or TI83+) 

First set the equation equal to zero:

Since the exponents are binomials, be sure to put parentheses around each of them, and write this equation in the form y1= for graphing purposes.  Graph this equation with the graphing calculator.  The standard window will be appropriate.

You can probably see that the answer this problem is the root (zero) that occurs at x = 8.  However, you can’t be sure from looking at it that the answer is exactly 8.  Also, most of the time, the answer will NOT come out even for these types of problems, so it may be helpful to go ahead and find the zero.  With a TI 83/84, use the keystrokes [2^nd]  [CALC F4]  and look for [2: zero] as illustrated below. 

You will be asked for a “left bound”, so give the calculator a value to the left of what you think the zero is, let’s say any number between zero and 7 inclusive, and press [ENTER].  Then the calculator will want a “right bound”, so give it a number larger than 8, say a 9 or a 10 and press [ENTER].  Then the calculator says “Guess?”, so press [ENTER] again, and the following screen appears:

This means that the zero or root is at x = 8.  Don’t worry about the fact that it says y = 230.  You are working with an exponential graph, which is UNBELIEVABLY steep.  On this calculator one pixel (that is one DOT!) of movement to the right moves about .21 units to the right or left, but it moves a vertical distance of somewhere around 4.3 x10¹³ up or down.  This is close as the calculator can come to y = 0. 

Final Answer:  x= 8

P.  523 # 16.

Solution:         

Take the ln of both sides of the equation:

Using the second law of logarithms:

Using the distributive property:

Get all x terms on the left side by subtracting    from each side of the equation:

Get the x variable in one place by factoring out the common factor of  x:

Solve for x by dividing each side of the equation by  

In the final step, parentheses are included that would be suitable for calculating the answer with a TI 83 or TI 84.  To calculate with a TI 85 or TI 86, use parentheses as indicated in the previous step.

Calculate with your calculator:

                        x   ≈  -4.64385618978 . . . or -4.64

You may also want to use the graphing calculator to obtain the approximate solution to this exercise by setting the equation  equal to zero, and graphing  .   Never mind what the graph looks like, just use the “zeros” function [2^nd]  [F4 (Calc)]  [2 (zeros)]  of the TI 83 or TI 84 or the “ROOT” method of the TI 85 or TI 86 to find the x-intercept of that graph.  It will amazingly enough be exactly the same (at least to calculator accuracy) as the answer given above!  Please see the graph below that was contributed by Department Chairman Lane Vosbury.

 :

P. 523 # 17. 

Solution:       

Take the LN of both sides of the equation:

Using the second law of logarithms:

Using the distributive property:

Get all x terms on the left side by subtracting   from each side of the equation and get the non-x terms on the right side by adding :

Get the x variable in one place by factoring out the common factor of  x:

Solve for x by dividing each side of the equation by

In the final step, parentheses are included that would be suitable for calculating the answer with a TI 83 or TI 84.  To calculate with a TI 85 or TI 86, use parentheses as indicated in the previous step.

Calculate with your calculator:

                         x ≈ −2.451964916 . . .     which rounds off to −2.45

P. 523 # 19. 

Solution:       

Take the LN of both sides of the equation:

Using the second law of logarithms:

Using the distributive property:

Get all x terms on the left side by subtracting   from each side of the equation and get the non-x terms on the right side by adding :

Get the x variable in one place by factoring out the common factor of  x:

Solve for x by dividing each side of the equation by

In the final step, parentheses are included that would be suitable for calculating the answer with a TI 83 or TI 84.  To calculate with a TI 85 or TI 86, use parentheses as indicated in the previous step.

Calculate with your calculator:  

Note:  It came out even!!!   Could this be a “contrived” problem??  Could it be solved as in the first few problems in this section using the laws of exponents??  Hmmm . . .

P. 525 # 23. 

Solution:  Since all the logarithms are already on the same side of the equation, you should begin by combining the logarithm terms on the left side into a single logarithm, using the first law of logarithms.  Remember, when you have a sum of logarithms, you get the log of a product.

Now convert this logarithmic form to exponential form, using the basic definition of logarithms.

                            means the same as 

Solve as a quadratic equation.

This factors into:

Of these answers,  must be rejected, since it results in a log of a negative in the original problem.  The final answer is .  

Extra Problem from algebra.com.

Solution:  Since all the logarithms are already on the same side of the equation, you should begin by combining the logarithm terms on the left side into a single logarithm, using the first law of logarithms.  Remember, when you have a sum of logarithms, you get the log of a product.

Now convert this logarithmic form to exponential form, using the basic definition of logarithms.

                             means the same as 

Solve as a quadratic equation.

This factors into:

Of these answers,  must be rejected, since it results in a log of a negative in the original problem.  The final answer is .             

Also it checks!  You can substitute  into the original equation:

P. 526 # 27. 

Solution:  You should begin by getting all the logarithm terms on one side, and any non-log terms on the other side.  In this case, subtract  from each side of the equation.  This leaves:

By the second law of logarithms, the difference of two logarithms can be expressed as a quotient:

Now convert this logarithmic form to exponential form, using the basic definition of logarithms.

                         means the same as 

Cross multiply, or multiply both sides of the equation by the LCD—it gives you the same result:

Solve as any linear equation.

This answer is acceptable, since when substituted into the original equation, it always results in the logs of positive numbers.

P. 527 # 29. 

Solution:       You should begin by getting all the logarithm terms on one side, and any non-log terms on the other side.  In this case, subtract  from each side of the equation.  This leaves:

By the second law of logarithms, the difference of two logarithms can be expressed as a quotient:

Now convert this logarithmic form to exponential form, using the basic definition of logarithms.

                              means the same as 

Cross multiply, or multiply both sides of the equation by the LCD—it gives you the same result:

Solve as any linear equation.

P. 528 # 33. 

Solution:     In the previous example, you were instructed to get all the log terms on one side and the non-log terms on the other side.  However, in this case, there are no non-log terms, so there is no need to do this step.  It is a good idea in this case to use the second law of logarithms to make the log of a quotient on the left.                      

Next, remember the “This Equals That” Theorem that was mentioned in #3?  Well, “This Equals That” applies to logarithms as well as exponents:

“This Equals That Theorem”:

       If   ,  then     

Now, you  may either cross multiply, or multiply both sides of the equation by the LCD—it gives you the same result:

Solve as a quadratic equation.

Set the equation equal to zero:

Factor the trinomial:

Set each factor equal to zero:

Now, you must check these answers by substituting them back into the original equation to make sure that neither answer gives you a log of a negative.  As it turns out, BOTH of these answers result in logs of negatives, so neither answer is acceptable.  Both must be rejected!  Therefore the final answer is NO SOLUTION!

P. 528 # 34. 

Solution:      As in the previous example, there are no non-log terms in the equation, so it will not be necessary to get all the log terms on one side.  It is a good idea in this case to use the second law of logarithms to make the log of a quotient on the left and also on the right side of the equation. 

Next, remember the “This Equals That” Theorem that was mentioned in the previous exercise?  Well, “This Equals That” is back again!!

“This Equals That Theorem”:

      If   ,  then        

Now, you  may either cross multiply, or multiply both sides of the equation by the LCD—it gives you the same result:

Solve as a quadratic equation.

Set the equation equal to zero:

Factor the trinomial:

Set each factor equal to zero:

Both answers are acceptable, since neither answer results in a log of a negative.

P. 528 # 35. 

Solution:     As in the previous examples, there are no non-log terms, so there is no need to get all the logs on one side of the equation.  It is a good idea in this case to use the first law of logarithms to make the log of a product on the left and also on the right side of the equation. 

Next, remember the “This Equals That” Theorem from the previous exercise?  Well, here it is again: 

“This Equals That Theorem”:

                               If   ,  then        

Solve as a quadratic equation, and set equal to zero.

Factor the trinomial:

Set each factor equal to zero:

The first answer, is acceptable, but you must reject the  since it results in a log of a negative.

P. 528 # 36. 

Solution:     As in the previous examples, there are no non-log terms, so there is no need to get all the logs on one side of the equation.  It is a good idea in this case to use the first law of logarithms to make the log of a product on the left and also on the right side of the equation. 

Next, remember the “This Equals That” Theorem from the previous exercises?  Well, here it is again: 

“This Equals That Theorem”:

      If   ,  then     

Solve as a quadratic equation, and set equal to zero.

Factor the trinomial:

Set each factor equal to zero:

The first answer must rejected since it results in a log of a negative (actually it results in two negatives, as it most always does!).  However, second answer  is acceptable.

Extra Problem from Keith.

Find the exact value, and also calculate the answer to the nearest millionth.

Solution:  This really means  which means                   

Now solve for x:                           

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