How to solve: $20\cdot x\cdot\log_2 x=1000000$

Question

$$20\cdot x\cdot\log_2 x=1000000$$ I tried but it is pretty strange to me. Thanks for your help!

Answer 1

The best you'll likely get is a solution involving the Lambert W function (also known as the product-log function). This function is typically examined in the situation in which we want to solve $xe^x = c$ for $x$, with constant $c$.

We begin by converting the $\log_2(x)$ into base $e$:

$$\log_2(x) = \frac{\ln(x)}{\ln(2)}$$

Then we solve for $x \ln(x)$ in your equation after substituting this:

$$x \ln(x) = 50,000 \cdot \ln(2)$$

We take the W function of both sides of this. This is nice since we have a property of the W function: $W(a \ln(a)) = \ln(a)$. Then

$$W(x \ln(x)) = \ln(x) = W(50,000 \cdot \ln(2))$$

Take the exponential, with base $e$, of both sides to solve for $x$:

$$x = e^{W(50,000 \cdot \ln(2))}$$

Typically from here you have to use some sort of numerical methods to approximate the solution. WolframAlpha gives the approximation

$$x \approx 4159.030805482706592671138392365184845680401116526152667975$$

Answer 2

You can only approximate it. But if you set $x = 2^k$ then

$x \log_2 x = 2^k*k = 50000$.

$\log_2 50000 = \frac {\ln 50000}{\ln 2} \approx 15.6$

$2^{15.6}*1 \approx 50000$

$2^{15.6}*2 = 2^{13.6}*4=2^{12.6}*8=2^{11.6}*16$. so to have $2^k*k=5000$ is to have $11.6 < k < 12.6$.

$2^{12}*12 = 49152$ which is close.

If we try a little closer we have $2^{12.1}*12.1 \approx 53118$.

So we know $2^{12}= 4096< x < 2^{12.1}\approx 4390$.

Using a calculator. we can find $2^{12.02}*12.02 \approx 49921$ and $2^{12.03}*12.03\approx 50310$ so $2^{12.02} \approx 4153 < x < 2^{12.03} \approx 4182$.

Still using a calculator I find $4159\log_2 4159 \approx 49999.6$ and $4160\log_2 4160 \approx 5013$.

So $x \approx 4159$ is about as close as you will get.

But read up on the Lambert W function.

The basic idea is that $x \log_2 x$ is an invertable function (as $\log_2 x$ is increasing and continuous and $x$ is as well. $x \log_2 x $ is monotonically increasing and continuous so for every $K\in \mathbb R$ there will be an $x$ so that $x \log_2 x = K$.

But there is no way of doing in algebraically. You can only approximate. (Which is actually the exact same thing with trig and log functions.)