Remember your laws of logs:
- $\log_{10}(XY) = \log_{10}(X)+\log_{10}(Y)$
- $\log_{10}(X/Y) = \log_{10}(X)-\log_{10}(Y)$
- $\log_{10}(X^Y) = Y\cdot\log_{10}(X)$
So for the first one: $\log_{10}(A)+2\cdot\log_{10}(1/A) $ $=\log_{10}(A)+2\cdot\log_{10}(A^{-1})$ $=\log_{10}(A)-2\cdot\log_{10}(A)$ $= - \log_{10}(A)=-a$
Alternatively: $\log_{10}(A)+2\cdot\log_{10}(1/A) $ $= \log_{10}(A)+2\cdot\log_{10}(1)-2\cdot\log_{10}(A) = a + 2\cdot 0 -2a = -a $ (because $\log(1)=0$)
For the third one, keep in mind $\sqrt[3]{C} = C^{1/3}$
Edit: The third...
Keep in mind the cube root is also in the denominator, so that argument is being divided as well and should have a minus sign in front of it. Next, $\log_{10}(100)=\log_{10}(10^2) = 2$.
$\log_{10}((100A^2)/(B^4 \cdot \sqrt[3]{C}))$ $=\log_{10}(100)+2\log_{10}(A)-4\log_{10}(B)-(1/3)\log_{10}(C)$ $=2+2a-4b-(1/3)c$