Prove that $\log _5 7 < \sqrt 2.$

Question

Prove that $\log _5 7 < \sqrt 2.$

Trial : Here $\log _5 7 < \sqrt 2 \implies 5^\sqrt 2 <7.$ But I don't know how to prove this. Please help.

Answer 1

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.

Answer 2

$f(x)=x^\frac1x$ 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 $$5^\sqrt2>5^{1.4}>7$$

Answer 3

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

• $7 < 5^{1.4} <5^{\sqrt{2}}$

Answer 4

Define $f(x)=5^{x/5}-x$.

If $x>5$ is obvious that $f'(x)=5^{(-1+x/5)}\ln (5)-1>0.$

Since $f(5)=0$, we have $f(7)>0$.

Since $\frac{7}{5}=\sqrt{\frac{49}{25}}<\sqrt{2}$ we are done.

Answer 5

Prove that $\log _5 7 < \sqrt 2$

Lemma

Mathematicians use the word "lemma" to refer to a fact about math which helps us prove other facts about math.

The following is a useful lemma:

Let $x$ and $y$ be any two positive decimal numbers.

For example,

• $x$ could be $\pi \approx 3.141592653589793238462643383282 \cdots$
• $x$ could be $\frac{4}{3}$
• $x$ could be the number $9$
• $x$ could be $0.407129911$

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  If
      there exists a positive whole number $k$ such that $\lfloor{x * 10^{k}}\rfloor > \lfloor{y * 10^{k}}\rfloor$
  then
     $x > y$

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If you are unfamiliar with the notation $\lfloor{x}\rfloor$, it means "delete everything to the right of the decimal point."

For example,

• $\lfloor{3.1459}\rfloor = 3$
• $\lfloor{45.28}\rfloor = 45$
• $\lfloor{1.5}\rfloor = 1$

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In other words, if you multiply both decimal numbers by ten a few times, and then throw away everything to the right of the decimal point, then you can sometimes tell which number is bigger.

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$ \begin{align} log_5(7) & \approx 1.209061 \dots \\ \sqrt{2} & \approx 1.4142136 \dots \\ \end{align} $

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$\begin{align} \lfloor 10*\sqrt{2} \rfloor & = 14 \\ \lfloor 10*\log_5(7)\rfloor & = 12 \\ \end{align} $

$14 = \lfloor 10*\sqrt{2} \rfloor > \lfloor 10*\log_5(7) \rfloor = 12$

Therefore, $\sqrt{2} > log_5{7}$