Log problem: log_{sec(theta)}log_3(n root of 3) = 2

[bobrossu]

Problem:
Find all values of n for which log_sec(theta)log₃(n root of 3) = 2 ,assume sec(theta)>0

I'm assuming the logs are multiplying. The first log only has a base though, so I am not sure. Is it possible it is log of a log ?

 

[Harry_the_cat]

Problem:
Find all values of n for which log_sec(theta)log₃(n root of 3) = 2 ,assume sec(theta)>0

I'm assuming the logs are multiplying. The first log only has a base though, so I am not sure. Is it possible it is log of a log ?

Yes it seems to be the log of a log. Should have a second set of brackets.

This leads to:

\(\displaystyle \log_{3}(3^{\frac{1}{n}}) = \sec^2(\theta)\)

Can you go from there?

 

[bobrossu]

Yes it seems to be the log of a log. Should have a second set of brackets.

This leads to:

\(\displaystyle \log_{3}(3^{\frac{1}{n}}) = \sec^2(\theta)\)

Can you go from there?

yea, thanks.
I think I got it...

1.) change
\(\displaystyle \log_{3}(3^{\frac{1}{n}}) = \sec^2(\theta)\) into ------> 3^{sec^2(theta)}=3^(1/n) -------> sec²(theta)=(1/n) -----> (1/cos²(theta))=(1/n) .... since sec(theta) > 0 then cos²(theta) or n must be 0 (<) n (</=) 1.