Melody

Wouldn't   f^-1(x)  have to be given a valid domain in order for it to be a function?

So    f( f^-1(x) ) =  f(x)  would only be strictly true for a given domain ?

But yes,  I guess that is what they are after.

\text{For how many values of n with } 0\le n\le 100 \text{ is the graph of}  f(x) = \sin \left(x + n\right) \text{ identical to the graph of } g(x) = \cos x?

\(\text{For how many values of n with } 0\le n\le 100 \text{ is the graph of}\\ f(x) = \sin \left(x + n\right) \text{ identical to the graph of } g(x) = \cos x? \)

\( 0\le n\le 100 \\ 0\le n\le 31.83\pi \\ \)

\(sin0=cos(0+\pi/2)\quad \text{then it will work every 2pi after that}\\ n=\frac{\pi}{2},\;\;\frac{5\pi}{2},\;\;\frac{9\pi}{2},\;\;.....\frac{61\pi}{2}\\ 1,5,9,4c-3, 4*16-3\\ \text{so that appears to be 16 times.} \)

Here is the graph

I have answered.

Here is a video on it

You are missing a bracket.

I do not know what you are trying to ask. 

Does the 7 refer to the 7th root?

I'm curious too.  

Let me assume base 10

\(2+log(2a)=3+log(3b)+log[6(a+b)]\\ log2+log a=1+log3+logb + log6 + log(a+b) \\ log2 - log3 - log6 -1= logb + log(a+b) - log a \\ log2 - log3 - log6 -log10= log(a+b) - log a + logb \\ log\frac{2}{3*6*10} = log\left[\frac{(a+b)}{a} *b\right] \\ \frac{2}{3*6*10} = \frac{(a+b)}{a} *b\\~\\ \frac{(a+b)}{a} *b=\frac{1}{90} \)