I have recently helped out some students work through issues involving exponential equations.  The equations causing grief are those in which the manipulated variable “x” makes more than ONE appearance; that is to say, those representing systems of equations involving two exponential functions.   I will provide several examples of such equations along with various perspectives on solving them. 

Through the process of creating this post, I decided to be very specific at various junctures so that the material contained within could be used by students themselves as they learn about exponential functions and equations.  It is not my intention to insult the intelligence of other math instructors who may read this article.

 

Prior to showing these examples, however, a short analysis of three simple exponential functions and their relationship with one another would be helpful.

 

Consider the functions below:

f(x) = 2^x , displayed with red in Graph 1 below

g(x) = 3^x , displayed with blue in Graph 1 below

h(x) = 6^x , displayed with black in Graph 1 below

 

 

Graph 1 

 

 

 

 

Also shown in the illustration above is the line x = 2; this vertical line intersects each of the exponential functions at the points labeled A (2, 4), B (2, 9) and C (2, 36).  That is to say the following:

f(2) = 4  à  A (2, 4)

g(2) = 9  à B (2, 9)

h(2) = 36  à  C (2, 36)

 

It is mportant to note here is that the product of “f(2)” and “g(2)” equals the value of “h(2)”; this will hold true for all values of “x”. 

 

As such, the following statement can be made for “h(x)”:

h(x) =(2^x)(3^x)  à  h(x) = 6^x

 

This notion is first introduced to students in junior high math as a “law of exponents”; showing how the graphs of the three functions are related with one another provides a valuable visual representation of why this “law” holds true. 

Stated in obvious terms, there should be no surprise that if you double something that has already tripled, the result will be an increase by factor of "6" over what existed at the outset. 

 

The functions ”f(x) = 2^x” and “g(x) = 3^x” must be “aligned” horizontally before representing their product as the simplified function “h(x) = 6^x”; that is to say, the functions being multiplied (or divided) must be raised to the same exponent in order to be “blended” to represent a third, more simplified, function.

 

This alignment places the two functions on the same "schedule" and also reveals the initial value of each function at a common starting position; the line "x = 0" is my preference here. 

 

 

Consider a small variation on the original scenario above. 

 

This time, the function “g(x) = 3^x” has been modified slightly by introducing a small phase shift as illustrated below.  Graph 2 illustrates this new relationship.

 

f(x) = 2^x , displayed with red in Graph 1 below

g(x) = 3^x+1 , displayed with blue in Graph 1 below

h(x) =(2^x)(3^x+1) , displayed with black in Graph 1 below

 

 

Graph 2 

 

 

 

This time, as the illustration above shows, g(2) = 27; therefore the product of the functions “f(x) = 2^x” and “g(x) = 3^x+1” at “x = 2” should equal the value of “h(x)” at x = 2.  A quick visual check on the graph above verifies this.

 

f(2) = 4  à  A (2, 4)

g(2) = 27  à  B (2, 27)

h(2) = 108  à  C (2, 108)

The goal with this line of reasoning is to “adjust” the appearance of our function “g(x) = 3^x+1” so that it “aligns” horizontally with the function “f(x) = 2^x”.  This is easily accomplished by either applying the “product law” of exponents or having a quick look at the graph of “g(x) = 3^x+1” above (making connections between the two is always favorable).  Either way, when “aligned” horizontally with the function “f(x) = 2^x”, the function “g(x) = 3^x+1” can be written as “g(x) = 3(3^x)”.  This function is shown with greater detail in the illustration below.

 

Graph 3 

 

 

 

Our function “g(x)”, as displayed in the illustration above, can take on various appearances depending on the “start point” chosen. 

 

The points A (-1, 1), B (0, 3), C (1, 9) and D (2, 27) are labeled on the function above, all of which could be used as a “start point”; the phase shift must account for the start point selected.

Shown below are four versions of the function “g(x)”, all of which represent the same function illustrated in Graph 3 above.

Start Point                                           Defining Function

A (-1, 1)……………………………………... g(x) = 1(3)^x+1

B (0, 3)……………………..…………..…... g(x) = 3(3)^x

C (1, 9)…………………………………..…... g(x) = 9(3)^x-1

D (2, 27)……………………………….....…. g(x) = 27(3)^x-2

All versions of the function “g(x)” are equivalent; the one chosen should fit the situation at hand (these variations are very powerful when working with geometric sequences as well). 

 

The version of function “g(x)” that will work to our advantage in the context set out at the beginning is the one whose “start point” resides on the “y-axis”, since it is in alignment with the function “f(x) = 2^x”.

Now that the functions “f(x)” and “g(x)” have been “aligned” horizontally, their product can be simplified easily.  This allows for a much “cleaner” version of “h(x)”, the function representing the product of the first two functions.  These are summarized below.

f(x) = 2^x

g(x) = 3(3^x)

h(x) =3(6^x)

This little procedure allows for a very quick and effective “clean-up” of an otherwise messy situation.

 

 

Consider the exponential equation below.

6^x = 8(3)^x

This equation can be easily solved by finding the point of intersection of the following functions:

f(x)  = 6^x

g(x) = 8(3)^x

 

The illustration below shows the functions f(x) and g(x) along with the point of intersection, P (3, 216).

 

Graph 4 

 

 

Consider the same exponential equation once again here.

6^x = 8(3)^x

Since “6^x” and “3^x” are perfectly “aligned” horizontally (they have the same exponent), our exponential equation above can be simplified easily by dividing by “3^x”.  This simple maneuver results in a very basic exponential equation shown below.

2^x= 8

x = 3

Our solution here, “x = 3”, is consistent with the “x-value” determined using simultaneously equations shown in Graph 4 above.

 

 

The next example shows two functions that are not “aligned” horizontally.

6^x = 24(3)^x-1

A very quick manipulation of this results in the following equation.

6^x = 24(3)^x(3)^-1

2^x= 8

x = 3

 

This method of simplifying first using some basic operations is, by itself, pointless without connecting exponential functions to “real-world” scenarios; then again, the same can be said of solving these equations with graphing technology by modeling systems of equations.  Having said that, laying down some theory BEFORE, DURING and AFTER the incorporation of “real-world” scenarios is, nonetheless, still beneficial.

The fact remains that we don’t want to waste time on calculations BUT we still want students to learn how to think mathematically.  The approach above meets both of those requirements; simplify first using mathematical reasoning and then incorporate technology to finish it off, if needed. 

 

Many proponents of “discovery-based” learning will no doubt scoff at this approach; give it a try first and see what your students think.  There is still plenty of room left for students to discover "real-life" scenarios and make connections with the theory presented here.

 

Furthermore, as in “GESTALT”, when students know what to look for, their eyes are drawn to immediately to problem areas.  Part of my job is to help students prepare themselves to deal with these issues in meaningful ways.

 

The final matter I would like to address here is the approach text-books take in solving exponential equations.  I’ve included two examples, each of which will be solved using three approaches.  Method 1 shows graphing technology, Method 2 shows the approach illustrated in many of the text-books I’ve seen, and Method 3 shows a quick simplification before introducing logarithms (as shown above).

 

Example 1

Find the value of “x” to the nearest whole number given the following equation:

2^x = 4(3)^2-x

Method 1 (Graphing)

Let  f(x) = 2^x and g(x) = 4(3)^2-x be two functions representing the equation above.  Graphing these functions simultaneously and finding the point of intersection provides a very quick route to the solution as shown in Graph 5 below.

 

Graph 5 

 

 

Method 2 (Logarithms)

This seems to be the algebraic “method of choice” favored by all textbooks and is shown below.

The common logarithm is introduced immediately to both sides of our equation.

log (2)^x = log (4(3)^2-x)

Laws of logarithms (having been previously derived) are applied with the purpose of “freeing” the variable “x” from its location in the exponent.

(x) log 2 = log4 + (2-x) log 3

Some multiplication followed by appropriate manipulations shown below will ultimately result in a solution for “x”.

(x) log 2 + (x) log 3= log 4 + 2log 3

x (log2+log3) = log 4 + log 3²

x = ^{(log 4 + log 9)}/ _(log2+log3)

With scenarios asking for the solution “as an exact value”, we would be finished at this point.  Because of the numbers chosen in this example, however, the result above can be simplified further by using the “product law” of logarithms.  This simplification is shown below.

x = ^{(log 4 + log 9)}/ _(log2+log3)  à  x = ^log36/_log6  à  x = log₆36

x = log₆6²

x = 2

 

A far less cumbersome approach is shown below; logarithms are once again introduced, but only after the original equation has been simplified using the same thought processes described earlier.

 

Method 3 (Simplify First)

Our knowledge of “Laws of Exponents” is put to work immediately to simplify the original equation.

2^x = 4(3)^2-x

2^x = 4(3)² (1/ 3)^x

Multiplying both sides by “3^x” leads to a very quick simplification of the original equation, as shown below.

6^x = 36

At this point, log₆ can be applied to both sides to “unlock” the variable “x” from its “exponent location”; this maneuver is not needed in this case because of the number selection but will be required in general.

log₆ 6^x = log₆ 36

x = 2

 

Once students know what to look for here, this last method is extremely quick and clean.

 

The equation “6^x = 36” from above can be solved be recognizing “36” as “6²” and comparing the exponents to solve.  More often than not, however, obvious cases such as these are few and far between.  For this reason, logarithms are introduced to make quick work of an otherwise annoying “guess and check” procedure.

 

 

Example 2

Find the value of “x”, to the nearest hundredth given the following equation:

3(2)^x = (6)^x-2

Method 1 (Graphing)

Let  f(x) = 3(2)^x and g(x) = (6)^x-2 be two functions representing the equation above.  Graphing these functions simultaneously and finding the point of intersection provides a very quick route to the solution as shown in Graph 2 below.

 

Graph 6 

 

 

Method 2 (Logarithms)

This seems to be the algebraic “method of choice” favored by all textbooks and is shown below.

The common logarithm is introduced immediately to both sides of our equation.

log (3(2)^x) = log (6)^x-2

Laws of logarithms (having been previously derived) are applied with the purpose of “freeing” the variable “x” from its location in the exponent.

log 3 + (x) log 2 = (x-2) log 6

Some multiplication followed by appropriate manipulations shown below will ultimately result in a solution for “x”.

log 3 + (x) log 2 = (x) log 6 - 2 log 6

 (x) log 6 - (x) log 2 = log 3 + 2 log 6

x (log 6 – log 2) = log 3 + log 36

x = ^{(log 3 + log 36)}/ _{(log6 – log2)}

  x = ^log108/_log3

x = 4.26  

 

 

Method 3 (Simplify First)

Our knowledge of “Laws of Exponents” is once again put to work immediately to simplify the original equation.

3(2)^x = (6)^x-2

3(2)^x = (6)^x (6)^-2

Multiplying by 6² and dividing by 2^x on both sides of our equation leads to a very quick simplification of the original equation as shown below.

3^x = 108

At this point, log₁₀ (common logarithm) can be applied to both sides to “unlock” the variable “x” from its “exponent location”;  solving for “x” from this point is virtually immediate.

log 3^x = log 108

(x) log 3 = log 108

  x = ^log108/_log3

x = 4.26

 

I’ll include one more example below showing “Method 3” to further illustrate its ease of use.

4(2)^3x = 7^x-1

4(8)^x = 7^x(7)^-1

(⁷/₈)^x = 28

log (⁷/₈)^x = log 28

x = log 28 / log (7/8)

or

x = ^log28/_(log7-log8)

 

Depending on the requirement stated at the outset, the solution can either be left as shown above or technology can be used to quickly determine a decimal approximation.  If a decimal approximation is the goal, the procedure is picked up for the following:

x = log 28 / log (7/8)

x = -24.95

 

 

My assertion is straight forward.  Know what the solution means and how it can be represented on a graph; that being said, it is very often far quicker to simplify FIRST before calling in technology to the rescue.  As an added bonus, students may actually earn a deeper appreciation for the subject and feel empowered as a result.

 

The End