1

Can someone please explain to me how the last step in this simplification works.

$$2^x = x^2$$ $$\implies \ln(2^x) = \ln(x^2) \quad \forall x \ne 0 $$ $$\implies x \ln(2) = 2 \ln(x) $$ $$\implies \ln(x) = {2x \over \ln(2)} \quad \textbf {(A)}$$

From what I can tell the final result should be

$$\ln(x) = {x\ln(2) \over 2}$$

Robert Z
  • 145,942
ClownInTheMoon
  • 1,987

2 Answers2

1

To solve $x$, you need to use the product log function:

$$2^x=x^2\Longleftrightarrow x=-\frac{2\text{W}_n\left(\pm\frac{\ln(2)}{2}\right)}{\ln(2)}$$

Where $n\in\mathbb{Z}$.

And for the real solutions, we get three solutions:

  1. $$x_1=-\frac{2\text{W}\left(\frac{\ln(2)}{2}\right)}{\ln(2)}\approx-0.76666$$
  2. $$x_2=-\frac{2\text{W}_0\left(-\frac{\ln(2)}{2}\right)}{\ln(2)}=2$$
  3. $$x_3=-\frac{2\text{W}_{-1}\left(-\frac{\ln(2)}{2}\right)}{\ln(2)}=4$$

When $x\in\mathbb{R}$:

$$2^x=x^2\Longleftrightarrow\ln(2^x)=\ln(x^2)\Longleftrightarrow x\ln(2)=2\ln(x)\Longleftrightarrow\frac{\ln(2)}{2}=\frac{\ln(x)}{x}$$

Jan Eerland
  • 28,671
0

You're correct and their final step is upside down. I can't think why.

In regards to analytically solving the quation, a rule of thumb is that if you have an unknown in both a base and power, you'll need the Lambert $W$ function. If we search your equation in WolframAlpha, we get $x=-2\frac{W(\ln(2)/2)}{\ln(2)}$, where $W$ is a branch of the Lambert $W$ function.

Jam
  • 10,325