It's a layout... that I worked hard on. ^_^
----------------------------------------------------
Originally posted by Katzeal
I'm wrong >.>
Quite in fact.
Anyways... first one.
Turn that (1/x
1/2) into an (x
-1/2). Factor out an (x
-1/2) from the whole expression and you should be able to figure out what to do from there.
~~~~~
As for the trigonometric stuff...
Two and Three: The basic idea is to isolate the trigonometric function, and then solve. So for the first one, isolate the cos(x) turning it into...
cos(x) = -1/2
Now, just find a value on the unit circle that has a cosine that equals -1/2. In this case, it would be 2pi/3 and 4pi/3.
~~~~~
Four: This requires you to remember your trigonometric properties. (sin
2(x) + cos
2(x) = 1)
It should be easy to solve for that.
~~~~~
Five is all about logrithmic properties. In this case, remember that log
m(a) + log
m(b) = log
m(a * b).
~~~~~
Six is tricky. What we have here is an example of a hidden quadradic equation. So, if you expand it, you get (e
x - 2)(e
x - 5). Set both of the expressions in the parenthesis equal to 0 and solve like you would any normal equation. (For the record, ln(e) = 1, ln(e
x) = x)
~~~~~
Seven. Going back to log properties, ln = log
e, and log
m(a) - log
m(b) = log
m(a / b)
Hopefully this should put you on the right track...

____________________
Something goes here... eventually...