Consider the equality

$$a^x = bx + c$$

For simplicity and to avoid trivial cases, I would like to limit $a,b,c$ to be positive real numbers where $a \ne 1$ and $b \ne 0$. (Solving those two cases is easy.) I can't really see a way to manipulate this using logarithm and exponent properties.

How might one solve for $x$? Moreover, what if I have an inequality instead and want to solve for $x$ then?

(There is a bit of context in me asking this question and posting my own answer. Namely, I often see a fair number of questions in this vein on the site, albeit with specified constants or perhaps an inequality thrown in. This answer handles a lot of these through its generality, so I think there is merit in posting and answering this question, as it could serve as a useful place to direct more specific questions.)

  • 2
    $\begingroup$ This is indeed a very good idea ! So $2 \times (+1)$ $\endgroup$ Aug 14, 2020 at 9:52

1 Answer 1



To begin, whenever one finds an equation of this form to solve, they almost always have to resort to use of the Lambert W function - though a deceptively simple equation, it's actually very difficult to solve for $x$ when it is in these two different positions (as a linear term and in an exponent). Some simpler cases (such as $2^x = x+2$) can often be solved by guess-and-check, but otherwise the W function will be as close one gets to a closed form solution.

The Lambert W function $W(x)$ is the function which is the inverse to $f(x)=xe^x$. That is,

$$W(x)e^{W(x)} = x = W(xe^x)$$

Luckily this function is fairly well studied, and thus can be approximated, so you might have numerical solutions; I suggest looking at the linked Wikipedia article or elsewhere for ideas on doing so.

Note that the Lambert W function can be defined for complex numbers as well; throughout the below derivations, assume $W=W_0$, i.e. we are working on the principle branch.

General Case:

For now, however, we focus on handling the equality in its full generality as stated:

$$a^x = bx + c \tag 1$$

To start with, we'll make things easier for us: let $u := bx+c$, which gives $x = (u-c)/b$. Then $(1)$ becomes

$$a^{(u-c)/b} = u \tag 2$$

Our goal for now is to get $u$; we'll want to create something of the form $we^w$ on one side of the equation (where $w$ contains our variable of concern, $u$, and the variable only appears on that side). We'll divide both sides by the left-hand side; the negative exponent that results can be used to reverse the subtraction in the numerator. Then we get

$$ua^{(c-u)/b} = 1 \tag 3$$

We'll use some exponent properties next: note that

$$a^{(c-u)/b} = a^{c/b - u/b} = a^{c/b} a^{-u/b}$$

We apply this to $(3)$ and then divide both sides by $a^{c/b}$ to achieve this:

$$ua^{-u/b} = a^{-c/b} \tag 4$$

We want to continue making the form $we^w$ appear; at this point, it would be prudent to note the familiar property

$$x^y = e^{\log(x^y)} = e^{ y \log(x)}$$

We apply this in the case of the $a^{-u/b}$ term in $(4)$ to obtain

$$u \exp \left( -\frac u b \log(a) \right) = a^{-c/b} \tag 5$$

Now, clearly, if we multiply both sides by $-\log(a)/b$, we'll attain the desired form, so we do so:

$$-\frac u b \log(a) \exp \left(- \frac u b \log(a) \right) = -\frac{\log(a) a^{-c/b}}{b} \tag 6$$

Next, we apply the W function to both sides of $(6)$. Recall that $W(we^w) = w$; in our case, $w = u \log(a) / b$. The right-hand side won't be able to be simplified however. Regardless, we obtain

$$-\frac u b \log(a) = W \left( -\frac{\log(a) a^{-c/b}}{b} \right) \tag 7$$

Multiply both sides by $-b/\log(a)$; this lets us solve for $u$, and then we can bring $x$ back into the picture:

$$u = bx + c = -\frac{b}{\log(a)} W \left( -\frac{\log(a) a^{-c/b}}{b} \right) \tag 8$$

Solve for $x$ by subtracting $c$ and dividing by $b$ (note the latter gets rid of the $b$ in the coefficient of the W function):

$$x = - \frac c b -\frac{1}{\log(a)} W \left( -\frac{\log(a) a^{-c/b}}{b} \right) \tag 9$$

It would be prudent to use this general solution in a few equations alongside properties of the W function to show the use of the general solution above.

Example $\#1$:

Consider the equation

$$e^x = x + 1$$

On inspection, clearly $x=0$ is a solution, but we can verify this. Notice that, here, $a=e$ and $b=c=1$. Then, using the general form,

$$x = - \frac{1}{1} - \frac{1}{\log(e)} W \left( - \frac{\log(e)e^{-1/1}}{1} \right)$$

Thankfully, $\log(e) = 1$ and other simplifications occur, yielding

$$x = - 1 - W(-1/e)$$

This is a special value of the Lambert W function: $W(-1/e) = -1$. Thus, from this, we conclude $x=0$. If one would rather follow the same process as when we derived the general form, this is how it would go:

$$\begin{align} e^x = x + 1 &\implies e^{u-1} = u \\ &\implies ue^{1-u} = 1 \\ &\implies ue^1e^{-u} = 1 \\ &\implies ue^{-u} = 1/e \\ &\implies -ue^{-u} = - 1/e \\ &\implies W(-ue^{-u}) = W(-1/e) = -1 \\ &\implies -u = -1 \\ &\implies u = 1 \\ &\implies x+1 = 1 \\ &\implies x = 0 \end{align}$$

Example $\#2$:

Consider the equation

$$2^x = 2x$$

In this case, $a=b=2$ and $c=0$. $x=1$ is a clear solution on inspection, but let's verify. Then, per the general solution in $(9)$,

$$x = - \frac{0}{2} - \frac{1}{\log(2)} W \left( - \frac{\log(2)2^{0/2}}{2} \right) = - \frac{1}{\log(2)} W \left( - \frac{\log(2)}{2} \right)$$

Another useful property of the W function is that $W(-\log(a)/a) = -\log(a)$ (whenever $a \in [1/e,e]$). This lets us simplify the expression and negate the factor of $-1/\log(2)$, to conclude indeed $x=1$.

If one preferred to follow the more standard derivation without appealing to the general form:

$$\begin{align} 2^x = 2x &\implies 2^{u/2} = u \\ &\implies u2^{-u/2} = 1 \\ &\implies u\exp \left( - \frac u 2 \log(2) \right) = 1 \\ &\implies -\frac u 2 \log(2) \exp \left( - \frac u 2 \log(2) \right) = -\frac{\log(2)}{2} \\ &\implies W \left( -\frac u 2 \log(2) \exp \left( - \frac u 2 \log(2) \right) \right) = W \left( -\frac{\log(2)}{2} \right) \\ &\implies - \frac u 2 \log(2) = - \log(2) \\ &\implies \frac u 2 = 1 \\ &\implies \frac{2x}{2} = 1 \\ &\implies x = 1 \end{align}$$

Example $\#3$:

Consider the final example equation of

$$2^x = 3x + 4$$

We can only really appeal to the general form or process here, and not any special properties or values. Substitution of $a=2,b=3,c=4$ gives us

$$x = - \frac{4}{3} - \frac{1}{\log(2)} W \left( - \frac{\log(2)2^{-4/3}}{3} \right)$$

Wolfram gives an exact answer of $x=4$ (which would be for the principal branch), and an approximation of $x \approx -1.18692$ for the lower branch $W_{-1}$. (This is because the principal branch only covers $x \ge -1$; the $W_{-1}$ branch covers the remainder of the reals.) To derive the exact answer in the more typical manner,

$$\begin{align} 2^x = 3x + 4 &\implies 2^{(u-4)/3} = u \\ &\implies 2^{u/3} 2^{-4/3} = u \\ &\implies u2^{-u/3} = 2^{-4/3} \\ &\implies u \exp \left( - \frac u 3 \log(2) \right) = 2^{-4/3} \\ &\implies - \frac u 3 \log(2) \exp \left( - \frac u 3 \log(2) \right) = - \frac{\log(2)2^{-4/3}}{3} \\ &\implies - \frac u 3 \log(2) = W \left( - \frac{\log(2)2^{-4/3}}{3} \right) \\ &\implies u = - \frac{3}{\log(2)} W \left( - \frac{\log(2)2^{-4/3}}{3} \right) \\ &\implies 3x + 4 = - \frac{3}{\log(2)} W \left( - \frac{\log(2)2^{-4/3}}{3} \right) \\ &\implies x = - \frac 4 3 - \frac{1}{\log(2)} W \left( - \frac{\log(2)2^{-4/3}}{3} \right) \end{align}$$

(I'm honestly not sure how to manipulate this into $x=4$, however.)

Some Final Notes:

To handle the case of inequalities, at least on the principle branch, note that $W(x)$ is a monotonic-increasing function. Thus, if $x < y$, then $W(x)<W(y)$. Accordingly, for instance, if our equation in Example $\#3$ was instead $2^x > 3x+4$, we could claim, at the end,

$$x < - \frac 4 3 - \frac{1}{\log(2)} W \left( - \frac{\log(2)2^{-4/3}}{3} \right)$$

(Bear in mind two multiplications by $-1$ happen throughout the solving process, at $(6)$ and $(8)$ in the general form. At one point we also flip the order of equality, at $(3)$, since that's how these are normally displayed. Thus, there's a change in the inequality.)

Thus, to solve inequalities of the form $a^x > bx +c$ (or any other desired order), solve $a^x = bx + c$, and, once you have your solution of $x = f(a,b,c)$, then you can replace that equality with the opposite order (here, $x < f(a,b,c)$).

Finally, while not strictly the purpose of this question, I think it might be important to cover the exempted cases of $b= 0$ and $a=1$ as well...

Special Case $(b = 0)$:

If $b=0$, the previous process doesn't work; let's start at the beginning:

$$a^x = c$$

Take the logarithm of both sides, and you can solve for $x$ easily:

$$\log(a^x) = x \log(a) = \log(c) \implies x = \frac{\log(c)}{\log(a)}$$

Special Case $(a = 1)$:

In this case, our equation is

$$1^x = bx + c$$

Since $1^x = 1$ for all $x$, this is a trivial equation to solve, for which

$$x = \frac{1-c}b$$

Special Case $(b = 0 \land a = 1)$:

In this case our equation is $1^x = 0x + c$ - equivalent to $1=c$ - which is true for no $x$, unless $c=1$. Then it is true for all $x$.


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