Solving x⁴ ln(x + 1) = 1 Graphically

Grades 11-12

Question: y = x^4 ln(x + 1) y = 1

To find the solution to the system of equations

y = x^4 \ln(x + 1)

and

y = 1

, we look for the intersection point(s) of a quartic-logarithmic function and a horizontal line.

The function

f(x) = x^4 \ln(x + 1)

is defined only where the argument of the natural logarithm is positive, meaning

x + 1 > 0

, or

x > -1

Behavior for $-1 < x < 0$ : In this interval, $x^4$ is positive but $\ln(x + 1)$ is negative. Consequently, the entire expression is negative. Since we are looking for where the function equals $1$ (a positive value), there are no solutions in this range.
Behavior at $x = 0$ : The function evaluates to $0^4 \cdot \ln(1) = 0$ .
Behavior for $x > 0$ : Both $x^4$ and $\ln(x + 1)$ are positive and strictly increasing. This guarantees that their product is also strictly increasing for $x > 0$ . Because the function starts at $0$ (when $x=0$ ) and grows toward infinity, it must cross the horizontal line $y = 1$ exactly once.

In the diagram, the blue curve represents the function

y = x^4 \ln(x + 1)

. Notice how it drops toward negative infinity as

x

approaches the vertical asymptote at

x = -1

, and how it remains below the x-axis until

x=0

The horizontal red line represents

y = 1

. The intersection point

P

indicates the x-value that satisfies the equation. By observing the graph, we can see that this happens slightly to the right of

x = 1

If we test values:

At $x = 1$ : $y = 1^4 \ln(2) \approx 0.693$ . Since $0.693 < 1$ , the root is further to the right.
At $x = 1.1$ : $y = (1.1)^4 \ln(2.1) \approx 1.464 \cdot 0.742 \approx 1.08$ . The intersection occurs at approximately $x \approx 1.081$ .