Question

prove that log(n+1) to the base n > log(n+2) to the base (n+1)​

Answer 1

Answer:

Just simply look at the function f(t)=t−log(t). You can show that this function is always increasing and that f(n)≥f(1)=1 for every n.

Answer 2

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\: log_{n}(n + 1) > log_{n + 1}(n + 2)$

$\large\underline{\sf{Solution-}}$

Without loss of inequality, n > 1

We know,

$\rm :\longmapsto\:n < n + 1$

$\bf\implies \:\dfrac{1}{n} > \dfrac{1}{n + 1}$

$\bf\implies \:\dfrac{1}{n} + 1 > \dfrac{1}{n + 1} + 1$

$\bf\implies \:\dfrac{1 + n}{n} > \dfrac{1 + 1 + n}{n + 1}$

$\bf\implies \:\dfrac{1 + n}{n} > \dfrac{2+ n}{n + 1}$

Taking log to the base n on both sides, we get

$\rm :\longmapsto\: log_{n}\bigg[\dfrac{n + 1}{n}\bigg] > log_{n}\bigg[\dfrac{n + 2}{n + 1}\bigg] \: \: as \: n > 1$

Now,

Let assume that

$\rm :\longmapsto\: log_{n}(z) = x \: \: and \: log_{n + 1}(z) = y$

$\rm :\longmapsto\:z = {n}^{x} \: \: and \: \: z = {(n + 1)}^{y}$

$\bf\implies \: {n}^{x} = {(n + 1)}^{y}$

For equality to be maintained,

$\rm :\longmapsto\:x > y \: as \: n \: < \: n + 1$

$\rm :\longmapsto\: log_{n}(z) \: > \: log_{n + 1}(z)$

$\red{\rm :\longmapsto\:On \: replacing \: z \: by \: \dfrac{n + 2}{n + 1}}$

$\rm :\longmapsto\:log_{n}\bigg[\dfrac{n + 2}{n + 1}\bigg] > log_{n + 1}\bigg[\dfrac{n + 2}{n + 1}\bigg]$

As,

$\rm :\longmapsto\: log_{n}\bigg[\dfrac{n + 1}{n}\bigg] > log_{n}\bigg[\dfrac{n + 2}{n + 1}\bigg] \: \:$

$\bf\implies \:\: log_{n}\bigg[\dfrac{n + 1}{n}\bigg] > log_{n + 1}\bigg[\dfrac{n + 2}{n + 1}\bigg] \: \:$

We know that,

$\boxed{ \bf{ \: log_{a}\bigg[\dfrac{x}{y}\bigg] = log_{a}(x) - log_{a}(y)}}$

So, using this we get

$\rm :\longmapsto\: log_{n}(n + 1) - log_{n}(n) > log_{n + 1}(n + 2) - log_{n + 1}(n + 1)$

We know,

$\boxed{ \bf{ \: log_{x}(x) = 1}}$

So, using this, we get

$\rm :\longmapsto\: log_{n}(n + 1) - 1 > log_{n + 1}(n + 2) - 1$

$\rm :\longmapsto\: log_{n}(n + 1) > log_{n + 1}(n + 2)$

Hence, Proved

Additional Information :-

$\boxed{ \bf{ \: log(xy) = log(x) + log(y)}}$

$\boxed{ \bf{ \: log \frac{x}{y} = log(x) - log(y)}}$

$\boxed{ \bf{ \: log {x}^{y} = y \: logx}}$

$\boxed{ \bf{ \: {e}^{logx} = x}}$

$\boxed{ \bf{ \: {a}^{ log_{a}(x) } = x}}$

$\boxed{ \bf{ \: {a}^{ ylog_{a}(x) } = {x}^{y} }}$

$\boxed{ \bf{ \: log_{ {x}^{y} }( {x}^{z} ) = \frac{z}{y}}}$