Answer:
-2B
(I guess this is what you are looking for; didn't need A or C).
Step-by-step explanation:
It seems like to wants us to to find [tex]\log_9(\frac{1}{16})[/tex] in terms of [tex]A,B,C[/tex].
First thing I'm going to do is rewrite [tex]\log_9(\frac{1}{16})[/tex] using the quotient rule.
The quotient rule says:
[tex]\log_m(\frac{a}{b})=\log_m(a)-\log_m(b)[/tex]
So that means for our expression we have:
[tex]\log_9(\frac{1}{16})=\log_9(1)-\log_9(16)[/tex]
Second thing I'm going to do is say that [tex]\log_9(1)=0 \text{ since } 9^0=1[/tex].
[tex]\log_9(\frac{1}{16})=\log_9(1)-\log_9(16)[/tex]
[tex]\log_9(\frac{1}{16})=-\log_9(16)[/tex]
Now I know 16 is 4 squared so the third thing I'm going to do is replace 16 with 4^2 with aim to use power rule.
[tex]\log_9(\frac{1}{16})=\log_9(1)-\log_9(16)[/tex]
[tex]\log_9(\frac{1}{16})=-\log_9(16)[/tex]
[tex]\log_9(\frac{1}{16})=-\log_9(4^2)[/tex]
The fourth thing I'm going to is apply the power rule. The power rule say [tex]\log_a(b^x)=x\log_a(b)[/tex]. So I'm applying that now:
[tex]\log_9(\frac{1}{16})=\log_9(1)-\log_9(16)[/tex]
[tex]\log_9(\frac{1}{16})=-\log_9(16)[/tex]
[tex]\log_9(\frac{1}{16})=-2\log_9(4)[/tex]
So we are given that [tex]\log_9(4)[/tex] is [tex]B[/tex]. So this is the last thing I'm going to do is apply that substitution:
[tex]\log_9(\frac{1}{16})=\log_9(1)-\log_9(16)[/tex]
[tex]\log_9(\frac{1}{16})=-\log_9(16)[/tex]
[tex]\log_9(\frac{1}{16})=-2\log_9(4)[/tex]
[tex]\log_9(\frac{1}{16})=-2B[/tex]
