How can I compare logarithm and the number?

Question

There are two numbers $\log_3 4$ and $\sqrt[4]{2}$. How they can be compared without calculator?

Answer 1

It's not trivial in general case. But for your expression: $$ I=\frac{\sqrt[4]{2}}{\log_3 4}=\frac{a}{\log_3 a^{8}}=\frac18\frac{a}{\log_3 a}=f(a). $$

where $a=2^{1/4}$. Function $f(x)=x/(8\log_3x)=\frac{\ln 3}8 x/\ln x$ is decreasing from $1$ to $e$ (one can show it by taking derivative of $x/\ln x$).

Let's take $b=3^{2/13}$. First, we show that $b<a$ (it's not easy without calculator, but doable): $$ 3^{2/13} \quad ? \quad 2^{1/4}\\ 3^8\quad ? \quad2^{13}\\ 6561 \quad < \quad 8192$$ then we calculate $f(b)$: $$ f(b) = \frac{3^{2/13}}{8\log_33^{2/13}}= \frac{13}{16} 3^{2/13}. $$ Then we show that $f(b) < 1$ (still doable): $$ 13\cdot3^{2/13} \quad ?< \quad 16\\ 13^{13}3^2 \quad ?< \quad 16^{13}\\ 13^{12}3^2 \quad ?< \quad 16^{12}\\ 13^{6}3 \quad ?< \quad 16^{6}=2^{24}\\ \text{since}\ 13^23=507<512=2^9\\ 13^{4} \quad ?< \quad 2^{15}\qquad \\28561 \quad < \quad 32678 $$

Finally, $f(x)$ is decreasing means that if $a>b$, then $f(a) < f(b) < 1$.