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# P2 Jan 05 q5 Logs watch

1. 5. Find, giving your answer to 3 significant figures where appropriate, the value of x for which:

(c) ln sin x = –ln sec x, in the interval 0 < x < 90°.

sec x = 1/ cos x
sin x = cos x ⇒ tan x = 1 x = 45]

>>>just wondering how to get to the second line please?
2. If you *e* both sides, it gets rid of the Ln's

say you had lnx = 2, that would equal x = e^2
3. Ln sinx + Ln secx = 0 (bringing Ln secx to other side)

Ln (sinx * 1/cosx) = 0

Take exponent from both side

Sinx/Cosx = 1
Tanx = 1
x = 45

Please if you have the markscheme in electronic form, may you upload it somewhere or send it to me please :X =)
4. thanks both for the help..much appreciated!

(Original post by LT0918)
Please if you have the markscheme in electronic form, may you upload it somewhere or send it to me please :X =)
yeh i have the mark scheme..perhaps pm me ur email n i'l send it over!?
5. Received - thanks very much

For others the paper can be found here:

http://www.apxc57.dsl.pipex.com/04%2...k%20scheme.pdf

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