Prove that $\log _5 7 < \sqrt 2.$
Observe that: $$ \begin{align*} \log_5 7 &= \dfrac{3}{3}\log_5 7 \\ &= \dfrac{1}{3}\log_5 7^3 \\ &= \dfrac{1}{3}\log_5 343 \\ &< \dfrac{1}{3}\log_5 625\\ &= \dfrac{1}{3}\log_5 5^4\\ &= \dfrac{1}{3}(4)\\ &= \sqrt{\dfrac{16}{9}}\\ &< \sqrt{\dfrac{18}{9}}\\ &= \sqrt{2}\\ \end{align*} $$ as desired.
$f(x)=x^\frac1x$ is a function defined on $(0,\infty)$ its log is $$G(x)=\log f(x)=\frac{\log x}x$$ $$G'(x)=\frac{1-\log x}{x^2}$$ Therefore $G(x)=\log f(x)$ strictly decreases for $x>e$, but logarithm is monotone on $(0,\infty)$ so that $f(x)$ is strictly decreasing for $x>e$
This gives us $$5^{\frac15}>7^{\frac17}$$ implying (by taking 7th power) that $$5^\sqrt2>5^{1.4}>7$$
Want to prove that
- $\log_5{7} = \frac{\lg{7}}{\lg{5}} < \sqrt{2}$
Equivalently we can show that
- $\lg{7} < \lg{5}\times\sqrt{2}$
- $7 < 5^{\sqrt{2}}$
where $\lg$ is the base 2 logarithm. Notice that
- $5\times5^{\frac{2}{5}}= 5^{1.4} <5^{\sqrt{2}}$
So can we show that $\frac{7}{5} < 5^{\frac{2}{5}}$? Sure, since $7<8=32768^{\frac{1}{5}}<78125^{\frac{1}{5}}$. Hence
- $7 < 5^{1.4} <5^{\sqrt{2}}$