Prove that ${log}_{n} (n + 1) > {log}_{(n + 1)} (n + 2)$ for any natural number $n > 1$ .

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Solution

Since $\frac{n + 1}{n} = 1 + \frac{1}{n} > 1 + \frac{1}{n + 1} = \frac{n + 2}{n + 1}$
We have for $n > 1$
${log}_{n} (\frac{n + 1}{n}) > {log}_{n + 1} (\frac{n + 2}{n}) > {log}_{n + 1} (\frac{n + 2}{n + 1})$
Using quotient law of logarithms, we get
${log}_{n} (n + 1) - {log}_{n} n > {log}_{(n + 1)} (n + 2) - {log}_{(n + 1)} (n + 1)$
As we know that ${log}_{a} (a) = 1$ , the above expression reduces to
${log}_{n} (n + 1) - 1 > {log}_{n + 1} (n + 2) - 1$
$\Rightarrow {log}_{n} (n + 1) > {log}_{n + 1} (n + 2)$
$∴ {log}_{n} (n + 1) > {log}_{n + 1} (n + 2)$ is proved

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