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Fundamental Laws of Logarithms

If α, β and γ...

Question

If $α, β$ and $γ$ are three consecutive positive integers, then the value of ${log}_{β} (1 + α γ)$ is

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Solution

Let $α = β - 1$ and $γ = β + 1$
${log}_{β} (1 + α γ)$
$= {log}_{β} (1 + (β - 1) (β + 1))$
$= {log}_{β} (1 + β^{2} - 1)$
$= {log}_{β} β^{2}$
$= 2 {log}_{β} β = 2$

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