Can You Solve Without A Calculator?

We'll prove the upper bound first. Let $L = \log_2 3 + \log_3 4 + \log_4 5 + \log_5 6$ , so we want to prove that $L<6$ .

Let $y = \log_x (x+1)> 1$ for $x>1$ , then $x^y = x + 1\quad \Rightarrow\quad y \ln x = \ln(x+1)\quad \Rightarrow\quad \dfrac yx +{\ln x} \cdot {dy/dx} = \dfrac1{x+1} \quad \Rightarrow \quad \dfrac{dy}{dx} = \dfrac{\ln x}{1/(x+1) - y/x} < \dfrac{\ln x}{1/x- y/x}< 0.$

Thus $y$ is a decreasing function. Let $y = f(x)$ , then $L = f(2) + f(3) + f(4) + f(5) < f(2) + 3f(3)$ .

Because $f(2) = \log_2 3 < \log_2 4 = 2$ and $4^3 < 3^4\Rightarrow 4 < 3^{4/3} \Rightarrow f(3) = \log_3 4 < \frac43$ . Thus, $L < f(2) + 3f(3) < 2 + 3\left(\frac43\right) = 6$ .

We'll prove that $L > 5$ by reversing engineering the result found by applying the arithmetic mean geometric mean inequality . That is, we want to prove that $L > 4\sqrt[4]{ f(2) \cdot f(3) \cdot f(4) \cdot f(5) } = 4\sqrt[4]{\log_2 6}> 5$ .

Claim: $4\sqrt[4]{\log_2 6}> 5$ .

Proof: Proving this claim is analogous to proving that $6 > 2^{(5/4)^4}$ , or equivalently $6^{256} > 2^{625} \Leftrightarrow 3^{256} > 2^{369}$ . This can be proven if we reverse engineer the inequality $3^2 > 2^3 \quad \Leftrightarrow \quad (3^2)^{125} > (2^3)^{125} \quad \Leftrightarrow \quad 3^{250} > 2^{375} \quad \Leftrightarrow \quad 3^{256} > 3^{250} > 2^{375} > 2^{369} .$

Hence, $5 < L < 6$ . Our answer is $\boxed 5$ .

[This is not a solution.]

Yes, if you type it into a calculator, you can get the value of 5.12 ish.

How can we demonstrate that

$5 < \log_2 3 + \log_3 4 + \log_4 5 + \log_5 6 < 6 ?$