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Question Number 169878 by mathlove last updated on 11/May/22

Answered by nikif99 last updated on 11/May/22

$$\frac{\cancel{\log 3}}{\log 2}\cdot \frac{\cancel{\log 4}}{\cancel{\log 3}}\cdot \frac{\cancel{\log 5}}{\cancel{\log 4}}\cdot ...\cdot \frac{\log (a+1)}{\cancel{\log a}}=5\Rightarrow$$$$\frac{\log (a+1)}{\log 2}=5\Rightarrow \log (a+1)=5\log 2\Rightarrow$$$$\log (a+1)=\log 2^5\Rightarrow a+1=2^5\Rightarrow a=31$$

Answered by som(math1967) last updated on 11/May/22

$${log}_{2}3.{log}_{3}4.{log}_{4}5\bullet ...\bullet {log}_a(a+1)=5$$$$\Rightarrow {log}_{2}4.{log}_{4}5\bullet ...\bullet {log}_a(a+1)=5$$$$\Rightarrow {log}_{2}5\bullet ...\bullet {log}_a(a+1)=5$$$$\Rightarrow {log}_2(a+1)=5$$$$\Rightarrow a+1=2^5\therefore a=31$$

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