How to solve $\log_2 x = x-1$

Question

$$y=\log_2x=x-1$$ I thought to write $$2^{(x-1)}=x$$ What should I do next?

Answer 1

Using Derivative Test

Given $\log_{2}(x) = x-1\;,$ Here $x>0$

As you write here $2^{x-1}=x\Rightarrow 2^x=2x$

Now Here we have to find roots of $f(x) =0\;,$ Where $f(x)=2^x-2x$

Now $f'(x)=2^x\ln(2)-2$ and $f''(x)=2^x\cdot (\ln 2)^2>0\;\forall x\in \mathbb{R}$

So $f''(x)=0$ has no real roots

So Using $\bf{LMVT}\;,$ We get $f'(x)=0$ has at most one real roots

and $f(x) = 0$ has at most $2$ real roots

So here we get $x=1$ and $x=2$ are the roots of $f(x)=0$

Answer 2

By the Bernoulli inequality $$ \forall x\not\in(1,2):\quad 2^{x-1}=(1+1)^{x-1}\geq 1+x-1=x $$ with equality precicely when $x=1$ or $x=2$, and $$ \forall x\in(1,2):\quad 2^{x-1}=(1+1)^{x-1}<1+x-1=x. $$ Hence, the only (real) solutions are $x=1$ and $x=2$.

Answer 3

This sort of equation in general requires a non-elementary function, the Lambert $W$ function. It is defined as the inverse of:

$$x=we^{w}$$

That is, given an $x$, find $w$. There are two branches when $-1<x<1/e$.

Now, starting with: $$2^{x-1}= x$$

We substitute: $$\frac{1}{2}e^{x\ln 2} = x$$

or $$-\frac{\ln 2}{2}= (-x\ln 2)e^{-x\ln 2}$$

Applying $W$, we get:

$$-x\ln 2 = W\left(-\frac{\ln 2}{2}\right)$$

Or:

$$x=\frac{-1}{\ln 2}W\left(-\frac{\ln 2}{2}\right)$$

Since $0<\ln 2/2<1/e$, we get two different values.

That doesn't explain why the values are $1$ and $2$, of course.

It is generally true that one of the values of $W\left(-\frac{\ln n}{n}\right)$ is $-\ln n$. This is equivalent to $1$ always being a solution to $n^{x-1}=x$.

The existence of another solution, when $n\neq e$, is because the graph of $n^{x-1}$ is greater than $x$ as $x\to +\infty$ and $x\to -\infty$. (When $n=e$, the curves are tangent at $x=1$.)

Answer 4

Well, $\log_2(x)$ is a concave function on $\mathbb{R}^+$ and $$ \log_2(x)=x-1 $$ clearly holds for $x=1$ and $x=2$. By concavity, it cannot hold for any other $x\in\mathbb{R}^+$.