Find $x$ such that $5^x - \sqrt{2x} -\log_2{x} = 22$

Question

Find $x$ such that $5^x - \sqrt{2x} -\log_2{x} = 22$.

I have observed that the solution of this equation should be $x = 2$, I also plotted the graph of the function $f(x) = 5^x - \sqrt{2x} -\log_2{x}$, and it looks like an increasing function, so the solution should be unique.

However my problem is proving that this solution is unique, and showing that the function is increasing using its derivative does not seem to work, as the expression is pretty ugly.

Do you have any suggestions on how to solve this without using the derivative?

Answer 1

Actually, you should write $$f(x)=5^x-\sqrt{2x}-\log_2(x)-22.$$ Now $$f'(x)=5^x\log(5)-\left(\frac{\sqrt{2}}{2\sqrt{x}}+\frac{1}{x\log(2)}\right).$$ Note that $$x\mapsto\frac{\sqrt{2}}{2\sqrt{x}}+\frac{1}{x\log(2)}$$ is decreasing on $[2,\infty)$, while $$x\mapsto 5^x\log(5)$$ is increasing on $[2,\infty)$. Hence, $f'(x)$ is increasing on $[2,\infty)$. A quick computation shows that $f'(2)>0$, and since $f'$ is increasing on $[2,\infty)$ that means $f'(x)>0$ for all $[2,\infty)$.

Answer 2

The function is a sum of three convex functions ($5^x$, $-\sqrt{2x}$ and $-\log_2 x$), hence the function is itself convex. Now the limit at zero is $\infty$ and $f(2)=22$. The derivative at $2$ is positive. Thus there should be exactly one more point in $]0,2[$ where the value is $22$, and no such point in $]2,\infty[$.

Answer 3

As saulspatz commented, the key point is that, since $x>0$ the second derivative is always positive. $$f(x)=5^x- \sqrt{2x}-\frac{\log (x)}{\log (2)}-22\implies f''(x)=\frac{1}{2 \sqrt{2} x^{3/2}}+\frac{1}{x^2 \log (2)}+5^x \log ^2(5)$$

Atking into account the behavior of $\log(x)$ close to $x=0^+$, there is another root. Expand $f(x)$ as a Taylor series at $x=0$ to get $$f(x)=-\left(\frac{\log (x)}{\log (2)}+21\right)-\sqrt{2} \sqrt{x}+x \log (5)+O\left(x^{3/2}\right)$$ Using the first term only $$\frac{\log (x)}{\log (2)}+21=0 \implies x=2^{-21}\approx 4.76837\times 10^{-7}$$ while solving the equation using Newton method with $x_0=2^{-21}$ would give the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 4.768371582\times 10^{-7} \\ 1 & 4.765147490 \times 10^{-7}\\ 2 & 4.765148580\times 10^{-7} \end{array} \right)$$