How do you solve this equation $ \log_{2}(x) = \sqrt x$? [closed]

Question

Disclaimer: Guys before voting to get the question closed I strongly feel we should instead have a feature on MSE that can merge such similar/duplicate questions since we got some really cool/through answers on both of them.

Context: Basically was needed in understanding How would you prove $\log_{2}x < \sqrt x$ for $x > 16$?. Please unlock this question and the discussions so I can paste them into this linked question (giving due credit ofc to the authors) so I can then ask my clarifying questions to one of the authors that I was trying to last nigh but those specific questions comments got locked for some reason?

I’m having a hard time solving the equation $\log_{2}(x) = \sqrt x$.

I tried squaring both sides and taking log of both sides, but it’s not really helping.

Kindly please help.

Answer 1

As you already wrote in the comments, $x=4$ and $x=16$ are both solutions, so this is just a proof that they are the only solutions.

Let's consider $f(x)=\sqrt{x} - \log_2 x$ function.

Because $$f'(x)=\frac{1}{2\sqrt{x}} - \frac{1}{x\ln 2}=\frac{\sqrt{x}\ln 2 - 2 }{2x \ln 2},$$ $f$ has a minimum at $x=\frac{4}{\ln^2 2}\approx 8.325$. From left of minimum on set $\left(0,\frac{4}{\ln^2 2}\right)$ $f$ is strictly decreasing, because there holds $f'(x)<0$, and on set $\left(\frac{4}{\ln^2 2},+\infty \right)$ strictly increasing, because there we have $f'(x)>0$. So, sign of derivative is enough for conclusion about minimum and we do not need second derivative.

On other hand we know $f(4)=f(16)=0$ and as $4<\frac{4}{\ln^2 2}<16$ then on set $(4,16)$ function $f$ is strictly negative and on set $(0,4)\cup(16,+\infty)$ strictly positive.

Answer 2

An alternative sledgehammer route is via the Lambert $W$ function, on both the principal branch $W_0$ and the other real branch $W_{-1}$.

$$ \log_2 x = \sqrt{x} $$ $$ 2\log_2 \sqrt{x} = \sqrt{x} $$ $$ 2\frac{\ln \sqrt{x}}{\ln 2} = \sqrt{x} $$ $$ \ln \sqrt{x} = \left(\frac12 \ln 2\right) \sqrt{x} $$ $$ \sqrt{x} = e^{\left(\frac12 \ln 2\right) \sqrt{x}} $$ $$ \sqrt{x} e^{\left(-\frac12 \ln 2\right) \sqrt{x}} = 1 $$ $$ \sqrt{x} e^{\left(\frac12 \ln \frac12\right) \sqrt{x}} = 1 $$ $$ \left(\frac12 \ln \frac12\right) \sqrt{x} e^{\left(\frac12 \ln \frac12\right) \sqrt{x}} = \left(\frac12 \ln \frac12\right) $$ $$ \left(\frac12 \ln \frac12\right) \sqrt{x} = W \left(\frac12 \ln \frac12\right) $$ $$ x = \left[ \frac{W \left(\frac12 \ln \frac12\right)}{\frac12 \ln \frac12} \right]^2 $$

(There's probably a faster way to get there, but I'm not the best Lambert manipulator.) Fortunately, we have an identity: When $z \geq 1/e, W_0(z \ln z) = \ln z$, so the principal branch $W(z) = W_0(z)$ gives us

$$ x = \left[ \frac{\ln \frac12}{\frac12 \ln \frac12} \right]^2 = 2^2 = 4 $$

What about the other value? That comes from the other branch, for which there is a corresponding identity: When $0 < z \leq 1/e, W_{-1}(z \ln z) = \ln z$. This would seem difficult to apply to the current equation, since $1/2 > 1/e$. However, note that we have the equality

$$ \frac12 \ln \frac12 = \frac14 \ln \frac14 $$

so we can now also obtain, from $W(z) = W_{-1}(z)$,

$$ x = \left[ \frac{W \left(\frac14 \ln \frac14\right)}{\frac14 \ln \frac14} \right]^2 $$ $$ x = \left[ \frac{\ln \frac14}{\frac14 \ln \frac14} \right]^2 = 4^2 = 16 $$

as desired.

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To be sure, there's a certain unsatisfying roundabout-ness to this approach, but it's another way of tackling it.