exponential equations with expansion...

Question

Solve for $x$ in the equation algebraically $$ 2^x=2x. $$

The solutions are $x =\{1,2\}.$ I have solved it but no one has validated my method. So I thought this website can help. I converted to logarithm, ended up with a binomial expansion and then evaluated limits of the series as it tends to zero and infinity. I don't know if it's right. I need help.

Answer 1

$$(2^x-2x)''=2^x\ln^22>0,$$ which says that the graph of $y=2^x-2x$ and a $x$-axes have two common points maximum.

Thus, by your work we got the answer: $\{1,2\}$.

Answer 2

Another method: $$ \begin{align} 2^x &= 2 x \iff \\ 1/2 &= x \, 2^{-x} = x \, e^{-\ln(2) x} \iff \\ -\ln(2)/2 &= -\ln(2) x \, e^{-\ln(2) x} \iff \\ W(-\ln(2)/2) &= -\ln(2) x \iff \\ x &= -\frac{1}{\ln(2)} W\left(- \frac{\ln(2)}{2} \right) \end{align} $$ where $W$ is the inverse to $f(x) = x e^x$, the Lambert W "function".

It is defined for negative arguments $x \ge -1/e$ and a double valued relation there.

So we get two solutions.

Answer 3

The equation is transcendental and cannot be solved analytically. Anyway we can isolate the root(s). Indeed, the derivative of $2^x-2x$ is $\log2\cdot2^x-2$ has a single root at $x^*=\log_2\dfrac 2{\log 2}\approx1.5287\cdots$, corresponding to an extremum.

Then $f(x^*)=2^{x^*}-2x^*=\dfrac2{\log2}-2\log_2\dfrac 2{\log 2}\approx-0.1721\cdots$ is a negative number, while $f(-\infty)=f(\infty)=\infty$. As the function is continuous, it has exactly two roots, on either side of $x^*$.

It turns out that if we try the closest integers, $1$ and $2$, we find these roots. But if instead we try $x^*\pm1$ (like could be done in an automated solution), we will find two positive values, showing that the roots lie in $(x^*-1,x^*)$ and $(x^*,x^*+1)$.