Cynthia Bell

2021-12-31

Show that $\mathrm{log}(x+1)-\mathrm{log}\left(x\right)<\frac{1}{x}$ for $x>0$

Shawn Kim

Beginner2022-01-01Added 25 answers

The function $\mathrm{log}(1+t)$ is strictly concave and therefore its graph stays under its tangent line at $0:$ for any $t\ne 0$ and $t>-1$ ,

$\mathrm{log}(1+t)<t.$

Your inequality is equivalent to

$\mathrm{log}(x+1)-\mathrm{log}\left(x\right)=\mathrm{log}(1+\frac{1}{x})<\frac{1}{x}$

Your inequality is equivalent to

Edward Patten

Beginner2022-01-02Added 38 answers

Consider $f\left(x\right)=\mathrm{log}\left(\frac{x+1}{x}\right)=\mathrm{log}(1+\frac{1}{x})$ . Set $u=\frac{1}{x}$ and then after making a good figure you will see that $\mathrm{log}(1+u)\le u$ and you are done.

Vasquez

Expert2022-01-09Added 669 answers

Let

and

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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A) No

B) 0

C) Yes

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B)$1.496\times {10}^{10}$

C)$1.496\times {10}^{12}$

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