Section 4.2: Properties of Logarithms

In the previous section, we introduced the logarithmic functions, log_b, as the inverses of the exponential functions exp_b(x) = b^x.  We saw that, although they cannot be given defined by any algebraic formula, they are needed to solve exponential equations.  In this section, we will discuss the algebraic properties of logarithmic functions, which we will use to calculate logarithms and solve exponential equations.  We will see that these properties follow directly from the rules of exponents.  We will begin, however, by discussing two very important logarithmic functions, log_e and log₁₀.

Two Special Logarithms

As you were reading the previous section, you may have been surprised to find that, despite the fact that there are many different logarithmic functions, your calculator has buttons for only two of them, namely, log_e and log₁₀.  In fact, you may not even have recognized them, since we traditional use a slightly different notation for these two, special logarithms.  Since we use a "base 10" number system (i.e., numbers are written in a way that implicitly uses powers of 10), it not surprising that we should treat the base 10 logarithms in a special way.  As we hinted at earlier, base 10 logarithms were first calculated by Henry Briggs in 1616, to simplify astronomical calculations.  Probably because this logarithm was used so commonly, people began to leave off the subscript.  This can be seen as early as 1647 in the writings of William Oughtred (Florian Cajori, A History of Mathematical Notations, vol. 1, p. 193).  That is,

If the base is not explicitly written, the base is assumed to be 10, so that log(x) = log₁₀(x).  This is sometimes referred to as the "common" logarithm.

Notice that values for this "common" logarithm are fairly easy to estimate to at least one correct digit.  For example, since log₁₀(2153) is the only solution to 10^x = 2153, and 10³ = 1000 < 2153 < 10000 = 10⁴, we know that 3 < log₁₀(2153) < 4.  Since the value in the middle of the first inequality was closer to the left than the right, we expect this to be true in the second inequality as well.  Therefore, we can estimate to one digit log₁₀(2153) » 3.  Similarly, since 10^-1 = 0.1 > 0.03125 > 0.01 = 10^-2, we know that -1 > log₁₀(0.03125) > -2, i.e., to one digit log₁₀(0.03125) » -2.  Note: If you are familiar with "scientific notation", that is, where we write 2153 = 2.153 × 10³ and 0.03125 = 3.125 × 10^-2, you can see that there is a close relationship between the value of the common logarithm of a number and its "order of magnitude" (i.e., the size of the exponent in scientific notation).

While the common logarithm is easier to compute, we have seen that the exponential function, e^x, with base e = 2.718..., arises naturally in a wide variety of applications, from banking to population growth.  Thus:

We call log_e the "natural" logarithm and abbreviate it as "ln".  That is, ln(x) = log_e(x).  

Although this logarithmic function was the first to be calculated, by John Napier in 1614, it was quickly superceded by the common logarithm in popular use.  It wasn't until Calculus has become widely used that mathematicians eventually realized that this was, in fact, the most "natural" logarithm to use.  The abbreviation, "ln", for the natural logarithm did not develop until even later.  One of the earliest known uses was only in 1893 (Florian Cajori, A History of Mathematical Notations, vol. 2, p. 107).  

You can see that these two logarithms (and their corresponding exponential functions) are automatically defined in XFunctions, in the menu on the left:

You should find those buttons on your calculator and learn how to use them.  In the next section, we will learn how to use them to compute values for every other logarithmic function.  

Make sure that you can use your calculator to compute values for log = log₁₀ and ln = log_e, by completing the following exercises.

Laws of Exponents and Logarithms

Since logarithmic and exponential functions are inverses, it is not surprising that the algebraic properties of exponential functions (expressed in the rules of exponents) should lead to useful algebraic formulas for logarithms.  Moreover, since all of the rules of exponents are reflected in symmetry properties of exponential graphs, we can begin to see the corresponding properties of logarithmic functions by looking at their graphs.

For example, we know that a horizontal shift in an exponential graph corresponds to a vertical scaling.  Since a logarithmic graph come from interchanging the horizontal and vertical axes in an exponential graph, this would suggest that a horizontal scaling in a logarithmic graph should correspond a vertical shift.  We can see this by comparing the graphs of, say, log₂(x) and log₂(8x), as in the Example in the following applet:

By looking at various points along the graph, such as:

you can see that log₂(8x) = log₂(x) + 3.  You may notice that the number "3" is somewhat special, since 8 = 2³.  That is, x = 3 is the solution to 8 = 2^x, or, using logarithm notation, log₂(8) = 3.  In particular, we have the equation: log₂(8x) = log₂(x) + log₂(8) = log₂(8) + log₂(x).  This suggests that logarithms satisfy the following rule:

Logarithms take products to sums.  That is, for any base, b > 0, and numbers x, y > 0, log_b(xy) = log_b(x) + log_b(y).

We can prove this very easily.  From rules of exponents, we know that b^{x + y} = b^xb^y, that is, exponential functions take a sum in the inputs to a product of the outputs.  Reversing the roles of inputs and outputs, this implies that logarithmic functions take product in the inputs to a sum of the outputs, which is exactly the rule of logarithms we have just stated.

If you would prefer a more algebraic proof, assign variable names to the values log_b(x) and log_b(y), say, u = log_b(x) and v = log_b(y).  By the definition of logarithms, we then know that b^u = x and b^v = y.  Multiplying the two equations, gives xy = b^ub^v = b^{u + v}.  Applying the definition of logarithms again gives log_b(xy) = u + v = log_b(x) + log_b(y), as expected.

As you might expect, this has a companion rule:

Logarithms take quotients to differences.  That is, for any base, b > 0, and numbers x, y > 0, log_b(x/y) = log_b(x) - log_b(y).

The proof is almost identical to that of the previous rule, so we will omit it.  However, this leads directly to another important rule.  Since exponential functions always take the value 1 for an input of 0, logarithmic functions always take the value 0 for an input of 1.  This means, if we take x = 1 in the quotient rule, that log_b(1/y) = log_b(1) - log_b(y) = 0 - log_b(y) = -log_b(y).  That is:

Logarithms take reciprocation to negation.  That is, for any base, b > 0, log_b(1/x) = -log_b(x).

We can easily derive a related rule.  If we let u = log_1/b(x), then (1/b)^u = x.  Taking reciprocals of both sides gives 1/x = b^u, so that log_b(1/x) = u = log_1/b(x).  That is.

For any base, b > 0, log_1/b(x) = log_b(1/x).

Taken together, these two formulas imply that:

For any base, b > 0, log_1/b(x) = log_b(1/x) = -log_b(x).  

In particular, a vertical flip of a logarithmic graph corresponds to taking a reciprocal of its base.  This corresponds to the fact, which we have already observed, that a horizontal flip in an exponential graph corresponds to taking a reciprocal of the base.  We can see this by comparing the graphs of, say, log₂(x) and log_1/2(x), as in the Example in the following applet:

Notice how the following points are on the graphs:

so that log_1/2(x) = -log₂(x).  

Our next example illustrates one of the most important properties of logarithms, as expressed by the following formula:

You can pull powers out of logarithms.  That is, for any base, b > 0, positive number, x > 0, and any number, y, log_b(x^y) = y log_b(x).

Notice that this includes the previous formula on reciprocals as a special case: with y = -1, this says that log_b(x^-1) = (-1)log_b(x); by rules of exponents, this is the same as, log_b(1/x) = -log_b(x).  In the next section, we will see how this property will allow us to directly relate exponential and linear models.  We will also show how this formula leads to the "change-of-base" formula, which will allow us to compute all logarithmic functions in terms of the natural and common logarithms (i.e., ln and log) on our calculators, and thus solve exponential equations.

We can verify this fact graphically, by comparing the graphs of, say, log₂(x) and log₂(x³), as in the Example in the following applet:

Notice how the following points are on the graphs:

so that log₂(x³) = 3log₂(x), as expected.  We can also prove this specific case, by using the product rule discussed earlier: log₂(x³) = log₂(xxx) = log₂(x) + log₂(x) + log₂(x) = 3log₂(x).  

The general, algebraic proof follows as before.  Let u = log_b(x) so that b^u = x.  Taking exponents of both sides gives x^y = (b^u)^y = b^uy = b^yu.  This implies that log_b(x^y) = yu = y log_b(x), as promised.  Notice how we use the rule of repeated exponents to prove this.  This means that, although it may not seem so at first, this rule of logarithms is the "reverse" of rule of repeated exponents.

We will now use this power formula to derive the most important formula of this section.  Remember how in the previous section, we suggested that log_k(x) could be written as ln(x)/ln(k).  We can show that this is only one case of the "change-of-base" formula:

Any two logarithmic functions are multiples of one another.  That is, for any bases, b and c > 0, there is a constant, K, so that log_b(x) = K log_c(x).  Specifically, log_b(x) = log_b(c) log_c(x); that is, K = log_b(c).  It is often more convenient to divide and write this as .

Since ln = log_e, we can now see that log_k(x) = ln(x)/ln(k) is an example of the latter formula with b = e and c = k.  

We can verify this directly, in the case of b = 4 and c = 2, by comparing the graphs of log₄ and log₂, as in the Example in the following applet:

Notice how the following points are on the graphs:

so that log₂(x) = 2 log₄(x) = log₂(4) log₄(x), since 2 = log₂(4) (i.e., 2² = 4).

The algebraic proof of this follows fairly easily from the previous rule of logarithms.  As usual, we begin by letting u = log_c(x) so that c^u = x.  Now apply log_b to both sides to obtain log_b(x) = log_b(c^u) = u log_b(c) = log_c(x)log_b(c) = log_b(c)log_c(x).  Notice how we used the previous rule of logarithms to obtain the second equality.  

While this may seem like a pretty esoteric property of logarithms, it is actually one of the very most important.  For instance, it justifies our statements about the relationship between logarithmic and exponential graphs of different bases.  More importantly, it says that the two buttons on our calculator are enough.  In fact, we need only one of them!  For example, we can calculate log₅(20) as log₁₀(20)/log₁₀(5) » (1.30103)/(0.69897) » 1.86135, which agrees to three digits with the value of 1.86 which we computed in the Exercises.  Note: In Calculus, and beyond, mathematicians tend rely almost exclusively on the natural logarithm, ln(x).  

Practice using this rule of logarithms by completing the following Exercises.  

Solving Exponential Equations

Armed with the change-of-base formula, we can now determine accurate solutions to many equations involving exponentials.  For some equations, we will need to use even more rules of logarithms.  However, we should emphasize that the basic process is the same as when doing algebra with simpler functions.  That is, we use the same rules of algebra, such as:

• Do the same thing to both sides of the equation.
• Work "outside-in", applying inverse functions, to "cancel" as much as possible.
• Substitute one equation into another.

It is just that the rules of exponents and rules of logarithms give us many new opportunities to make substitutions to "simplify" our equations.

For example, say we want to solve the equation:

-27 = -4·3^{(x - 2)/7} + 5

We should look at the functions which have been applied to the unknown, x:

• Subtract 2,
• Divide by 7,
• Apply the exponential with base 3,
• Apply the exponential with base 3,
• Multiply this by -4, and 
• Add 5 to the result.

As usual, we work "backwards" (i.e., "outside-in") to isolate x, by applying the corresponding inverse functions to both sides:

• Subtract 5 from both sides: -27 - 5 = -4·3^{(x - 2)/7} + 5 - 5, which gives -32 = -4·3^{(x - 2)/7}. 
• Divide both sides by -4: -32/(-4) = -4·3^{(x - 2)/7}/(-4), which gives 8 = 3^{(x - 2)/7}. 
• Apply the logarithm with base 3: 8 = 3^{(x - 2)/7}, which gives  (x - 2)/7 = log₃(8).
• Multiply by 7: 7(x - 2)/7 = 7 log₃(8), which gives  x - 2 = 7 log₃(8).
• Add 2: x - 2 + 2 = 7 log₃(8) + 2, which gives  x = 7log₃(8) + 2.

If we want a numerical value for the answer, we must use the change-of-base formula to write this as x = 7 ln(8)/ln(3) + 2.  Which we can plug into our calculator to obtain x » 15.2495.

This is a pretty typical example, when the unknown only appears once in an exponent.  However, when the unknown appears more than one place, we know that we must strive to rearrange and combine terms until the unknown finally appears in only one place.  Then we can finish as before.  For example, consider the equation:

2·5^{6(1 - x)} = 4·3^{(x - 2)/7}

We can't combine the x terms until we can get them out of the exponent.  Thankfully, the power rule for logarithms does just that for us.  While it works for any logarithm, we might want to choose either log₃ or log₅, since this will exactly cancel at least one of the exponential functions.  However, we must be careful to use this and other rules of logarithms properly:

Notice how, once we used the product, inverse, and power rules to bring x out of the exponent, the equations are simply linear equations (since the logarithms are just strange-looking constants).  That is why the last four steps are the same basic, algebraic steps we would use to solve a simpler-looking equation, like 2 + 6(1 - x) = 4 + ((x - 2)/7)3.

If we want to obtain a numerical answer, we must finally use the change-of-base formula, to obtain:

x = (log(2)/log(5) + 6 - log(4)/log(5) + 2log(3)/(7log(5)))/(log(3)/(7log(5)) + 6) » 0.945361.

We should mention that it is actually simpler if we choose to use log from the beginning.  Following the same steps:

we obtain a similar answer.  You can verify that these are the same results with a bit of algebra, or simply plug this into your calculator to obtain 0.945361.  You can also verify that this is a valid solution, since:

2·5^{6(1 - 0.971518)} » 3.38981 and 4·3^{(0.971518 - 2)/7} » 3.38981.

While these exponential equations are given out of context, in the next section, we will see how such equations arise naturally in a variety of real-world applications.

Before closing this section, we will stop to show how properties of logarithms can allow us to dispense with many of the other buttons on our calculator; in fact, those buttons probably work by using logarithms without you knowing it!  Note: Before the advent of hand-held digital calculators, detailed tables of values for the common and natural logarithms (with corresponding exponential tables) were the main tools used by mathematicians and scientists to compute products, quotients, powers, and roots.  

For example, assuming that our tables are accurate to 6 digits, if we were Sir Isaac Newton and we wanted to compute 134798 × 293146, we would rewrite this as: 

134798 × 293146 = 10^{log(134798 × 293146)} = 10^{log(134798) + log(293146)}
» 10^{5.12968 + 5.46708} » 10^10.5968 » 39518500000.

This is the same as the exact answer, 39515494508..., to 4 places.  Using logarithms, we were able to convert a hard multiplication problem into an easy addition problem, at the cost of three table look-ups and a slight loss of accuracy.

Similarly, we would compute a power, such as 3.14159^1.41421, as:

In this way, we convert the problem of computing a power into a multiplication problem, which then simplifies as:

10^{(1.41421)log(3.14159)} » 10^{(1.41421)(0.497150)} » 10^{(1.41421)(0.49715)}
» 10^0.703075 » 5.04748

This agrees with the exact value of 5.04747..., to 5 places.

Even with calculators, these are worthwhile techniques with which to be familiar.  We already have seen that to graph most logarithms in XFunctions (except for log₂, log₁₀, and ln), we must use the change-of-base formula to plot, say, log₅(x) as ln(x)/ln(5).  Likewise, to plot most powers and roots, such as 3^x or x^1/3, XFunctions actually uses the formula b^a = e^aln(b) to express these formulas as e^xln(3) and e^ln(x)/3.  That is why, if you try to plot the cube root function, f(x) = x^1/3, in XFunctions (using the formula "x^(1/3)), it will only show the graph for x > 0; while the domain of the cube root function is actually the entire number line, the domain of ln(x) and e^ln(x)/3 is only x > 0.  This is also why XFunctions pre-defines separate functions "sqrt" and "cubert" for the square and cube root functions.  Note: It defines these as piecewise functions, similar to the way we defined the absolute value function.

Note: Because mathematicians use the exponential with base e (i.e.,  exp_e = e^x), so often, it is common to drop the subscript and write this simply as "exp" (that is, while log = log₁₀, exp = exp_e); you can see that XFunctions uses this convention, as well, in its listing of pre-defined functions:

Practice using rules of logarithms by completing the following Exercises.  

Go to Exponential Models and Logarithms

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Send questions or comments to jwicks@northpark.edu

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