If x^x = 49, then x=?

In school mathematics, the unknown usually appears in one place, such as x^2=49 or 2x=49. But in x^x=49, x appears both as the base and the exponent, so standard algebra methods do not give a direct solution.

A useful method comes from equations of the form y e^y=a. Mathematicians including Leonhard Euler studied related equations in the 18th century. The function was later named the Lambert W function after Johann Heinrich Lambert.

Johann Heinrich Lambert

The Lambert W function is used to solve equations where the unknown appears both inside and outside an exponent. It can therefore be used to solve x^x=49. Without it, we generally find the solution using numerical methods such as trial and error or graphing rather than a standard algebraic formula.

Before solving the main equation, it helps to understand three ideas: the behavior of the function x^x, natural logarithms, and what the Lambert W function actually does.

The function x^x is always increasing (for x>0)

The expression x^x means "x multiplied by itself, x times" in a loose sense, though for non-whole numbers it is defined using exponentials. What matters here is its behavior: as x grows larger, x^x grows larger too, without ever turning back down. This is called an increasing function. For example,

33=27and44=2563^3 = 27\quad

\text{and}\quad

4^4 = 256

Since we want x^x = 49, and 27 < 49 < 256, we know the answer to our equation lies somewhere between x = 3 and x = 4.

Because x^x only increases and never decreases (for positive x), a horizontal line drawn at height 49 will cross the curve of x^x at exactly one point. This tells us something important: there is only one real positive solution to x^x = 49.

Natural logarithms

The natural logarithm, written ln, is a specific type of logarithm based on the number e (approximately 2.71828). Its most useful property, for our purposes, is called the power rule:

ln⁡(ab)=b⋅ln⁡(a)\ln(a^b) = b \cdot \ln(a)

The Lambert W function

If we have an expression that looks like:

a⋅eaa \cdot e^{a}

then applying the Lambert W function to it simply returns a:

W(a⋅ea)=aW(a \cdot e^{a}) = a

For example take the expression 3·e³, then,

W(3⋅e3)=3W(3 \cdot e^{3}) = 3

Solving x^x = 49

Now we put these three ideas together to solve the equation properly.

Step 1: Take the natural logarithm of both sides

Starting with:

xx=49x^x = 49

Apply ln to both sides:

ln⁡(xx)=ln⁡(49)\ln(x^x) = \ln(49)

Step 2: Apply the power rule

x⋅ln⁡(x)=ln⁡(49)x \cdot \ln(x) = \ln(49)

Step 3: Rewrite x in terms of e

Every positive number x can be written as e raised to the natural log of x, because ln and e "undo" each other:

x=eln⁡(x)x = e^{\ln(x)}

Substituting this into x·ln(x) = ln(49) gives:

eln⁡(x)⋅ln⁡(x)=ln⁡(49)e^{\ln(x)} \cdot \ln(x) = \ln(49)

We can reorder the multiplication,

ln⁡(x)⋅eln⁡(x)=ln⁡(49)\ln(x) \cdot e^{\ln(x)} = \ln(49)

Step 4: Apply the Lambert W function to both sides

W(ln⁡(x)⋅eln⁡(x))=W(ln⁡(49))W\big(\ln(x) \cdot e^{\ln(x)}\big) = W(\ln(49))

we get,

ln⁡(x)=W(ln⁡(49))\ln(x) = W(\ln(49))

At this point, there's no need to worry about how complicated the right side, W(ln(49)), looks. It is just a number — a constant we can calculate. The left side, ln(x), is also just an unknown constant we're trying to find. We are very close to separate x.

Step 5: Remove the logarithm

We have:

ln⁡(x)=W(ln⁡(49))\ln(x) = W(\ln(49))

To remove the natural logarithm, raise e to the power of both sides:

eln⁡(x)=eW(ln⁡(49))e^{\ln(x)} = e^{W(\ln(49))}

the left side simplifies:

x=eW(ln⁡(49))x = e^{W(\ln(49))}

This is the exact algebraic solution. The right-hand side may look unusual, but it is just a fixed number, calculable with any tool that supports the Lambert W function.

Step 6: Calculate the numeric value

Working out this expression numerically gives:

x≈3.278x \approx 3.278

This matches what we predicted earlier using the graph: the answer falls between 3 and 4.

Conclusion

In conclusion, xˣ = 49 cannot be solved using the usual methods used for equations such as x² = 49 because x appears both as the base and the exponent. Taking logarithms helps simplify the equation, but the Lambert W function is needed to obtain an exact form. The solution is x = eᵂ⁽ˡⁿ⁴⁹⁾, which is approximately 3.278. The same method can be used to solve equations of the form xˣ = k, where k is a known number.