Scared?

Let suppose: a=log20(log5(20)) b=log20(log20(5)) We know a=-b, therefore; F(a)+F(b)=[psin(a) + q×a^(1/2009)+15] + [psin(-a) + q×(-a^)(1/2009)+15]=30

$log_a(log_b(c))=-log_a(log_c(b))$

Therefore, if

• $a=log_{20}(log_5(20))$
• $b=log_{20}(log_{20}(5))$

Then

• $a=-b$ .
• $b=-a$ .

Plugging this into our function gives us $p(sin(a))+q(\sqrt[2009]{a})+15$ , and for the second part we get $p(sin(-a))+q(\sqrt[2009]{-a})+15$ .

Let's substitute

• $p(sin(a))=1$
• $q(\sqrt[2009]{a})=2$
• Note: We could substitute anything for these equations by manipulating $p$ and $q$ , I just chose these for simplicity.

Then, adding these functions together, we get

• $1+2+15-1-2+15$
• The 1 and 2 cancel out, leaving us with:
• $15+15=30$

Therefore, $\boxed{30}$ is our answer!

In conclusion- nothing to be scared of.

clearly ,if log20(log5(20))=a then we have to find f(a)+f(-a). then the odd functions will cancel out and the solution is 15+15=30

Assume the problem has a solution. Because Brilliant only allows numerical solutions, the answer must be independent of p and q. It must still hold, even if p and q are zero, in which case f(x) = 15, so the expression is 15+15=30.

log(base a)(log(base b)c) = -log(base a)(log(base c)b) because of log rules about the reciprocal, therefore power -1, being moved to the front as a multiplier instead.

Note log(base b)c = (log c)/(log b) any base

All this means that the value inside the second function is equal and opposite the value in the first.

Sine is an odd function so the results are equal and opposite and cancel when added.

2009th root also gives you equal and opposite answers to the two equal and opposite values and again cancels when added.

This leaves two 15's to add together.