0

$2^x=4x$

I cant seem to solve this equation. The furthest I have been able to come is $x-\log_2(x)=2$, but I can't figure how to solve. When I graph $2^x$ and $4x$ they intersect at $x=4$ and $x=0.31$, so I know it is possible to solve.

David G. Stork
  • 29,774
Bob the Turtle
  • 127
  • 5
    You can't solve this equation with standard algebra. –  Jan 13 '21 at 20:45
  • 5
    The second solution (near x=0.31) can't be expressed in elementary terms. It can be found using the Lambert-W function, but that's mostly a solution "by definition" rather than anything which clarifies the algebra. – Semiclassical Jan 13 '21 at 20:45
  • 3
    Just because a solution exists doesn't necessarily mean you can solve it using standard techniques. – DMcMor Jan 13 '21 at 20:46
  • 2
    $x=4$ seems to work. – John Douma Jan 13 '21 at 20:48
  • Check Wolfram, may help – Cooperation Jan 13 '21 at 20:52
  • @Semiclassical By the same token, the solution of $x^2=2$ near $1.41$ is also only a "solution by definition" rather than anything which clarifies the algebra. I know no sensible categorical distinction between using the square root in one case and using the Lambert $W$ function in the other. The only distinction is how often they are used, and that we mathematicians for subjective reasons have decided we like one more than the other. – Arthur Jan 13 '21 at 21:01

2 Answers2

4

The equation has indeed the easy integer solution $x=4$ which can be found by inspection. Graphing reveals a second solution (and it is possible to formally prove that there are no other real roots).

If you heard of the Taylor development, as the second root is not that large, you can develop $e^{x\ln2}$ to the second order and solve

$$1+x\ln2 +\frac{(x\ln2)^2}2=4x.$$

The smallest solution of this quadratic is $x=0.30935\cdots$, which is close the searched answer.

Even better, you can use the known approximation $0.31$ and write

$$e^{(0.31+y)\ln 2}=e^{0.31\ln 2}e^{y\ln 2}=4(0.31+y).$$

Solving as above,

$$y=-9.3067447651\cdots\,\cdot10^{-5}$$ and $$x=0.3099069323807\cdots,$$ where all decimals are exact !

-2

Classic problem:

$$ x = -\frac{W\left(-\frac{\log (2)}{4}\right)}{\log (2)}, {\rm or}\ -\frac{W_{-1}\left(-\frac{\log (2)}{4}\right)}{\log (2)}$$

where $W$ is the ProductLog function.

Numerical evaluation of these analytic solutions: $.309907$ and $4$.

enter image description here

David G. Stork
  • 29,774