A Generalized Logarithm for
Exponential-Linear Equations
Dan Kalman

Dan Kalman (kalman@email.cas.american.edu) joined the mathematics faculty at American University in 1993, following an eight year stint in the aerospace industry and earlier teaching positions in Wisconsin and South Dakota.
He has won three MAA writing awards, is an Associate
Editor of Mathematics Magazine, and served a term as
Associate Executive Director of the MAA. His interests include matrix algebra, curriculum development, and interactive computer environments for exploring mathematics, especially using Mathwright software.

How do you solve the equation
1.6x = 5054.4 − 122.35x?

(1)

We will refer to equations of this type, with an exponential expression on one side and a linear one on the other, as exponential-linear equations. Numerical approaches such as Newton’s method or bisection quickly lead to accurate approximate solutions of exponential-linear equations. But in terms of the elementary functions of calculus and college algebra, there is no analytic solution.
One approach to remedying this situation is to introduce a special function designed to solve exponential-linear equations. Quadratic equations, by way of analogy, are
√
solvable in terms of the special function x, which in turn is simply the inverse of a very special and simple quadratic function. Similarly, exponential equations are solvable in terms of the natural logarithm log, and that too is the inverse of a very special function. So it is reasonable to ask whether there is a special function in terms of which exponential-linear equations might be solved. Furthermore, an obvious strategy for finding such a function is to invert some simple function connected with exponentiallinear equations.
This line of thinking proves to be immediately successful, and leads to a function I call glog (pronounced gee-log) which is a kind of generalized logarithm. As intended, glog can be used to solve exponential-linear equations. But that is by no means all it is good for. For example,. with glog you can write a closed-form expression for the
.
x.

iterated exponential (x x ), and solve x + y = x y for y. The glog function is also closely related to another special function, called the Lambert W function in [3] and
[6], whose study dates to work of Lambert in 1758 and of Euler in 1777. Interesting questions about glog arise at every turn, from symbolic integration, to inequalities and estimation, to numerical computation. Elaborating these points is the goal of this paper.
Definition of glog
As indicated in the introduction, glog is defined as the inverse of a simple function, namely, e x /x. This definition is prompted by considering an especially simple type of exponential-linear equation, of the form e x = cx
2

(2)

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y = glog+ x y = log x

3
2
1

y = glog− x y = glog x −5

e
−1
e , −1

5

10

−1
−2

Figure 1. Graphs of y = glog x and y = log x

where c is an arbitrary constant. Rewriting this equation as ex =c x provokes our interest in inverting e x /x, leading to the following definition of glog: y = glog(x) iff x = e y /y iff e y = x y.

(3)

Two different graphical representations provide insight about glog. First, by the usual method of reflection in the line y = x, a graph of glog is easily obtained, as shown in Figure 1. For later reference, the log function is also included in the graph.
The graph reveals at once some of the gross features of glog. For one thing, glog is not a function because for x > e, it has two positive values. For x < 0 glog is well defined, and negative, but it is not defined for 0 ≤ x < e. When we need to distinguish between glog’s two positive values, we will call the larger glog+ and smaller glog− .
As suggested by the graph, glog− (x) ≤ log(x) ≤ glog+ (x) for all x ≥ e, with equality of all three expressions when x = e.
These inequalities appear obvious from the graph, but of course they can be derived analytically. For example, suppose y = glog x is greater than 1. By definition, e y /y = x so e y = x y > x. This shows that y > log x.
The second graphical representation depends on the fact that glog c is defined as the solution to e x = cx. The solutions to this equation can be visualized as the xcoordinates of the points where y = e x and y = cx intersect (Figure 2). That is, glog c is determined by the intersections of a line of slope c with the exponential curve.
This viewpoint provides additional insight about where glog is defined or single valued. When c < 0 the line has a negative slope and there is a unique intersection with the exponential curve. Lines with small positive slopes do not intersect the exponential curve at all, corresponding to values of c with no glog. There is a least positive slope at which line and curve intersect, that intersection being a point of tangency. It is easy to see that this least slope is at c = e with x = 1. For greater slopes there are two solutions to (2).
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8
6

c=e c>e 4 c y k+1 /(k + 1)!.
This implies that e y /y > y k /(k + 1)!. Now let y = glog+ (x). Then e y /y = x and y k = [glog+ (x)]k , so the previous inequality becomes x> [glog+ (x)]k
.
(k + 1)!

This in turn leads to k (k + 1)!x > glog+ (x).

This inequality can be used to show that, like the logarithm, the larger branch of glog
√
grows more slowly than any root function. That is, for any k, glog+ (x)/ k x → 0 as x → ∞.
Although this provides information about the growth rate, it is not of much use in estimating values of the glog function, because it relies on a very crude estimate: e y > y k /k!. A better estimate can be derived using e y > ay k , with a as large as possible.
This requires a point of tangency between f (y) = ay k and g(y) = e y , as illustrated in
Figure 3.
That is easy to arrange: simply demand that f (y) = g(y) and f (y) = g (y) hold simultaneously. In other words, solve the system ay k = e y aky k−1 = e y .
Clearly, these equations hold only if y = k and a = ek /k k . That means that e y ≥
(e/k)k y k with equality just at y = k. And because of the tangency condition, e y is very close to (e/k)k y k for y near k.

g(y) = e y

f (y) = ay k y Figure 3. Point of Tangency

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Arguing as before, we can now derive an estimate for glog. First, divide the inequality by y. e y /y ≥ (e/k)k y k−1
Take y = glog+ (x), so e y /y = x and y k−1 = [glog+ (x)]k−1 . x ≥ (e/k)k [glog+ (x)]k−1
Finally, solve for glog. k e

k−1

k

x ≥ glog+ (x)

(10)

with equality at x = ek /k. This inequality again bounds glog+ using a root function, but it also provides a very good estimate near x = ek /k. As a particular case, taking k = 2 leads to
2
e

glog+ (x) ≤

2

x

which implies glog+ (x) < x.
This is not apparent in Figure 1 due to dissimilar scales for the two axes.
Differentiation and Integration
To a college mathematics teacher, the urge to differentiate and integrate any new function that shows up on the scene is nearly irresistible. Although a thorough discussion will be too great a digression, these are topics that deserve at least a brief consideration.
See [8] for a more complete treatment.
Differentiating glog follows the standard pattern for differentiating inverse functions. Suppose y = glog x. Then x = e y /y, or e y = x y. Differentiate both sides with respect to x and solve for y , producing y =

ey

y
.
−x

But we already know that e y = x y, so y =

y
.
xy − x

Thus, the formula for the derivative of glog is glog (x) =

glog(x)
.
x glog(x) − x

In terms of the standard elementary functions of analysis, the glog is not integrable in closed form (see [10, Example 23]). However, since glog is itself not an elementary function (see [9]), it is natural to contemplate a larger class of functions, made up of
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combinations of glog and the usual elementary functions. Considered in the context of this larger class of functions, I do not know whether glog has an integral in closed form.
If there is such an integral, it can be of neither the form f (glog(x)) nor h(x, glog(x)) with f and h elementary functions.
Some other functions involving glog do have simple integrals, including glogn (x) for n > 1 and glogn (x)/x for n ≥ 1. Integrating these examples is left for the amusement of the reader.
Before leaving this topic, it is worth mentioning that the W function does have a simple integral: x(W (x) − 1 + 1/W (x)) ([6]). Consequently, W permits the integration of a number of differential equations, accounting for several of the applications of
W cited in [6]. It may be that glog, too, can be applied to solve differential equations that arise in a natural way, but that must remain a question for future investigation.

Computation of glog
Included in [6] is a thorough discussion of efficient accurate computation of W , for both real and complex values. By virtue of (9), this provides methods for computing glog as well, but it is beyond the scope of this paper to delve so deeply into these issues. However, I will state some of the computational results relevant to computing real values of glog.
Before proceeding, it will be helpful to partition the graph of glog into several segments, as indicated in Figure 4. In the figure, segments A and C are characterized by large values of |x| and small values of | glog(x)|. Conversely, on segment B, |x| takes on small values while | glog(x)| grows without bound. On segment E both x and glog(x) increase to infinity. Finally, segment D is a neighborhood of the branch point
(e, 1) where glog+ and glog− coincide.
On segments A and C, the expression
∞

glog(x) =

n=1

n n−1 n! 1 x n

converges for |x| > 1/e. This result follows from a similar series given for W in [6], which is based on the Lagrange inversion formula.
4
E

3
2
D

1
−5

C e A

5

10

−1
B
−2
Figure 4. Partitioned graph of glog.

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The connection of glog with inverting x/ log x in [2] was mentioned earlier. Specifically, y = x/ log x if and only if x = eglog y . Now [2] presents an expansion for this x in terms of y, L y = log y, and L 2 y = log log y : x = y(L y + L 2 y) +
+

L2 y y L2 y y L2 y
(2 − L 2 y) +
(6 − 9L 2 y + 2(L 2 y)2 )
+
2
Ly
2(L y)
6(Ly)3

y L2 y
(24 − 8L 2 y + 44(L 2 y)2 − 6(L 2 y)3 ) + · · · .
24(L y)4

Since log x = glog y this provides a means for computing glog. More specifically, it is easily inferred from the discussion that this expansion computes glog+ and hence corresponds to segment E of the graph. The accompanying text does not provide an error analysis or discussion of convergence, merely referring to work of Comtet ([5]) and Berg ([1]).
Citing another work of Comtet ([4]), [6] presents a similar expansion for W :
W (y) = L y − L 2 y +
+

L2 y
L2 y
L2 y
(L 2 y − 2) +
(6 − 9L 2 y + 2(L 2 y)2 )
+
Ly
2(L y)2
6(Ly)3

L2 y
(−24 + 72L 2 y − 44(L 2 y)2 + 6(L 2 y)3 ) + O
24(L y)4

L2 y
Ly

5

.

This expansion converges for y > e. With the identity glog x = −W (−1/x), we can use the above expansion to estimate glog x for −1/e < x < 0, with greatest accuracy nearest 0, corresponding to segment B of the graph.
The preceding remarks refer to every segment of the graph of glog except segment D. For that segment, we can use a second order Taylor approximation for the exponential function to derive an estimate of glog. By definition, y = glog x is equivalent to x = e y /y. The second order Taylor expansion for e y about y = 1 is .5e(y 2 + 1) with error around (y − 1)3 /3 for y near 1. Now substitute this approximation in the equation for x : x = (.5e)

y2 + 1
.
y

This equation can be solved as a quadratic in y, producing the estimate
1
glog x = y ≈ (x ± e x 2 − e2 )

for x ≥ e.
The results above provide glog estimates for each segment of the graph. These estimates may be refined through the use of Newton’s method. To evaluate glog a, the equation et = at must be solved for t. This is equivalent to solving the equation f a (t) = 0 where f a (t) = et − at.
Clearly, f a only has a root if a is in the domain of glog, ie., if a < 0 or a ≥ e.
Depending on which of these conditions holds, the behavior of f a takes one of two forms, as illustrated in Figure 5. For a < 0, f a is increasing on the real line, with a unique root. For a ≥ e, f a has a global minimum at log(a), and is monotonic to either
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fa

fa

3
2

−2

2

1
−3

3

1 t −1

t

−1

1

1

2

−1 ae Figure 5. Typical graphs for f a

side of this minimum. In particular, f a as two roots, separated by log(a). For all values of a, f a is concave up over the entire line.
These characteristics imply that Newton’s method is extremely well behaved for all of the f a . From any initial value (except log a in the case a ≥ e), Newton’s method must converge to a root of f a , and this convergence will be monotonic from at least the second iteration on. To be more specific, if the initial guess t0 is uphill from the root (that is, f a (t0 ) > 0), then the next iterate will be between t0 and the root, and so all succeeding iterates will move monotonically closer to the root. If t0 is downhill from the root, then t1 will be uphill, and the successive iterates will again move monotonically toward the root. When a < 0, Newton’s method will find the unique root of f a no matter how t0 is defined. For a ≥ e, from any initial value greater than log(a) Newton’s method will converge to the greater root; from an initial value less than log(a) convergence will occur to the lesser root.
The Newton’s method iteration for f a takes a simple algebraic form. The general recursion is tn+1 = tn −

f a (tn ) f a (tn )

which becomes tn+1 =

tn − 1
1 − e−tn

after algebraic simplification.
Although Newton’s method is perfectly robust for f a , computationally it is desirable to select t0 as close as possible to the root. The estimates for glog presented earlier are useful in this regard. However there are other estimates that are accessible using only elementary methods available to calculus students. The analysis presented for segment D of the graph is the kind of thing I have in mind. Similar methods can be constructed for the other segments of the graph.
For example, on segments A and C, we have |y| very small. Accordingly, we can again estimate e y with a quadratic Taylor polynomial, this time expanded about 0. As in the earlier discussion, the equation x = e y /y becomes a quadratic equation which can be solved for x ≈ glog y.
On segments B and E, a different approach is required. For segment E, we can use inequality (10). This provides a good estimate for glog x near x = ek /k. An appropriate k can either be selected by trial and error, or by precomputing ek /k for the
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first several positive integers k. Alternatively, crudely estimating k as the greatest integer in ln x provides a decent estimate for k up to about 50. A similar analysis can be developed for segment B.
Using the elementary estimates just discussed to initiate Newton’s method, glog can be computed to high accuracy in just a few iterations. By running a large number of numerical experiments, I found approximate optimal transition points between the segments of the graph. The table below shows how the domain of glog was partitioned, and summarizes the convergence results for a large variety of values of x. Generally, convergence to about nine or ten decimal digits was observed within the number of iterations specified in the table. These results are based on haphazard experimentation; actual performance may vary.
Segment

a Domain

A

a < −.42

B

−.42 ≤ a < 0

C

a ≥ 3.4

t0 Formula
√
a − 1 + a 2 − 2a − 1 k −k

−(k e /|a|)

; k = log |a|
√
a − 1 − a 2 − 2a − 1
1/k+1

Iterations
3
2
2

D (y ≥ 1)

e