How would you prove $\log_{2}x < \sqrt x$ for $x > 16$? [closed]

Question

I'm not really showing how to prove this, since I tried finding the $x$-intercepts/zeros of $f(x) = \sqrt x - \log_{2} x$ , and see that $x = 4, 16$ work but inspection, but I'm not sure how to ensure those are all the zeros.

Once I find the zeros I think I can check the intervals between the zeros and undefined points and do the one-point test in each interval (since if it's positive or negative for one point in the interval then it's positive or negative for all points in the interval).

From there I could prove if it's positive for all $x > 16$. Kindly please help.

Answer 1

Define $$f(x) = \sqrt{x} - \log_2 x, \quad x \in (0,\infty). \tag{1}$$ Then $f$ has no discontinuities and is smooth, because $\sqrt{x}$ and $\log_2 x$ are continuous and smooth on this interval. The derivative is $$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{x \log 2} = \frac{\sqrt{x} \log 2 - 2}{2 x \log 2}. \tag{2}$$ (Note: for the purposes of this discussion, when the base of the logarithm is unspecified, it is taken to be the natural logarithm.) Consequently, $f$ has the unique critical point $$x = x_0 = \frac{4}{\log^2 2}. \tag{3}$$ As the numerator of $(2)$ is positive when $x > x_0$ and negative when $x < x_0$, it follows that $f$ is increasing on $x \in (x_0 , \infty)$, and decreasing when $x \in (0, x_0)$. Since $$f(x_0) = \frac{2}{\log 2} - 2 \log_2 \frac{2}{\log 2} \approx -0.172143 < 0, \tag{4}$$ and you have already located the zeroes $f(4) = 4(16) = 0$, it follows that these roots must be unique.

Answer 2

Let $y = \log_{2} \left( 16 x \right) - \sqrt {16 x}$ for all $x > 0$. It is easy to see that $y = \log_{2} \left( x \right) - 4 \left( \sqrt {x} - 1 \right)$. Take its derivative: $\text {d} y / \text {d} x = \log_{2} \left( e \right) / x - 2 \, / \sqrt {x}$, so $\text {d} y / \text {d} x \ge 0$ if and only if $\log_{2} \left( e \right) \ge 2 \sqrt {x}$. However, we know that for all $x > 1$, $2 \sqrt {x} > 2$; meanwhile, $e < 4$, so $\log_{2} \left( e \right) < 2$. In summary, $\log_{2} \left( e \right) < 2 \sqrt {x}$ for all $x > 1$.

So $\text {d} y / \text {d} x < 0$ for all $x > 1$. Now $y = 0$ at $x = 1$, we swiftly conclude that $y < 0$ for all $x > 1$; that is, $\log_{2} \left( x \right) < \sqrt {x}$ for all $x > 16$. $\blacksquare$

Answer 3

First, observe that $e^2<9<16=2^4$ which implies $2<4\log 2$, also for $t\ge 16$ we get $$2\sqrt{t}<4\sqrt{t}\log 2$$ also, $4\le \sqrt t,\;$ so we get $$2\sqrt{t}<4\sqrt{t}\log 2\le t\log 2$$

then \begin{align*} \frac1{\log 2}\frac1t&<\dfrac12\dfrac1{\sqrt t}\quad \text{when } t\ge 16\\ \dfrac1{\log 2}\int_{16}^x\dfrac1t\,\mathrm dt&<\frac12\int_{16}^xt^{-1/2}\,\mathrm dt\quad \text{when } x> 16\\ \dfrac1{\log 2}\left(\log x-\log 16\right)&<\sqrt{x}-\sqrt{16}\\ \dfrac{\log x}{\log 2}-\dfrac{\log 16}{\log 2}&<\sqrt{x}-4\\ \log_2x-4&<\sqrt{x}-4 \end{align*}

Therefore $x>16\implies\log_2 x<\sqrt{x}$.

Answer 4

Let $x=2^{2t+2}.$ Then the inequality takes the form $2^t>t+1,$ $t>1.$ For $t=1$ we get the equality. So it suffices to show that $f(t)=2^t-t$ is increasing for $t\ge 1.$ We have $f'(t)=(\log 2)\, 2^t-1.$ Thus $f'(t)$ is increasing. Hence $$f'(t)>f'(1)=2\log 2 -1=\log 4-1>0,\quad t>1$$

Answer 5

Consider the function $$f(x) = \frac{\log_2(x)}{\sqrt{x}}$$ Taking the derivative, $$f'(x) = \frac{2 - \ln(x)}{x^{3/2} \ln(4)}$$ We see $f'(x) < 0$ for $x >e^2$, so $f(x)$ is decreasing for $x > e^2$.

Since $16 > e^2$ and $f(16) = 1$, $f(x) < 1$ for $x > 16$.

Answer 6

By @zkutch (and edited by @5xum):

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 and on set $\left(\frac{4}{\ln^2 2},+\infty \right)$ strictly increasing.

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.

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By @BrianTung:

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.