How to Solve $8n^2=64n\log_2(n)$

The logarithm is defined only when $n>0$, so $n=0$ is not a possible solution. For positive $n$, divide the equation by $8n$ and solve the simpler equation that remains. It does not have a simple integer solution, but it has two positive real solutions that can be found numerically.

Reduce the equation

Start with

$$8n^2=64n\log_2(n).$$

Since $n>0$, division by $8n$ is valid:

$$n=8\log_2(n).$$

This is a transcendental equation: the variable appears both outside and inside a logarithm. In general, you cannot solve it using only the usual algebraic operations and logarithm rules. First determine how many solutions exist, then approximate them.

Show that there are two positive solutions

Define a function whose zeros are the solutions:

$$f(n)=n-8\log_2(n),\qquad n>0.$$

The equation is equivalent to $f(n)=0$. Using $\log_2(n)=\ln(n)/\ln(2)$, differentiate:

$$f'(n)=1-\frac{8}{n\ln(2)}.$$

The derivative is zero at

$$n=\frac{8}{\ln(2)}\approx 11.542.$$

For smaller positive $n$, $f'(n)<0$; for larger $n$, $f'(n)>0$. Thus $f$ decreases to one minimum and then increases. Also, $f(n)$ tends to positive infinity as $n$ approaches zero from above, and it tends to positive infinity as $n$ grows without bound. The minimum is negative, so the graph crosses zero twice.

For example, $f(1)=1>0$, while $f(2)=2-8=-6<0$. This confirms a crossing on the left side of the minimum. On the right, $f(43.5)<0$ and $f(43.6)>0$, confirming another crossing. Because there is only one minimum, these are the only two roots.

Approximate the solutions

A calculator or numerical solver can find roots of $f(n)=0$. Newton’s method is one option. Starting with an estimate $n_k$, update it using

$$n_{k+1}=n_k-\frac{n_k-8\log_2(n_k)}{1-8/(n_k\ln(2))}.$$

Use an initial estimate near $1.1$ for the smaller root and one near $43.5$ for the larger root. The results are

$$n\approx 1.099997\quad\text{or}\quad n\approx 43.5592604.$$

Rounded to four decimal places, the solutions are $1.1000$ and $43.5593$. Substituting either value into $n=8\log_2(n)$ gives equality up to rounding error. The two solutions are easy to miss if you only test values near one of the roots.

Exact form using the Lambert W function

There is an exact expression using the Lambert W function, defined by $W(x)e^{W(x)}=x$. Rewrite the equation as

$$\ln(n)=\frac{\ln(2)}{8}n.$$

Rearranging gives $-(\ln(2)/8)n$ multiplied by its exponential equal to $-\ln(2)/8$. Therefore,

$$n=-\frac{8}{\ln(2)}W_k\!\left(-\frac{\ln(2)}{8}\right).$$

The argument lies between $-1/e$ and $0$, where the Lambert W function has two real branches. The branch $k=0$ gives the smaller root, and $k=-1$ gives the larger one. This notation is useful for an exact formula, but numerical approximations are usually more practical in an algebra or precalculus setting.

Common mistakes

• Keeping $n=0$ as a candidate: The original logarithm $\log_2(n)$ is undefined at zero.
• Dividing by $n$ without checking the domain: Here division by $n$ is safe only because the logarithm already requires $n>0$.
• Assuming there is only one answer: The function decreases and then increases, producing two positive roots.
• Treating a numerical approximation as an exact value: Values such as $1.1$ and $43.5593$ are rounded; they are not exact roots.
• Assuming an integer is required: Over the positive reals there are two solutions. If the problem requires integer $n$, neither root is an integer, so there are no integer solutions.