Let $a,b$, and $c$ be positive real numbers such that
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of
$$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1) ?$$
I tried to convert the entire thing to fractional logs and multiply the expression and add the two equations but it did not help.
Best Answer
As lab bhattacharjee commented, convert to natural logarithms and get $$\log_{a}b + \log_{b}c + \log_{c}a = 8\implies \frac{\log (b)}{\log (a)}+\frac{\log (a)}{\log (c)}+\frac{\log (c)}{\log (b)}=8$$ $$\log_{b}a + \log_{c}b + \log_{a}c = 13\implies \frac{\log (a)}{\log (b)}+\frac{\log (c)}{\log (a)}+\frac{\log (b)}{\log (c)}=13$$ Now expand $$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1)$$ which is $$\frac{\log (a)}{\log (b)}+\frac{\log (b)}{\log (a)}+\frac{\log (a)}{\log (c)}+\frac{\log (c)}{\log (a)}+\frac{\log (c)}{\log (b)}+\frac{\log (b)}{\log (c)}+2$$ I am sure that you can take it from here and finish.