Use the property of the difference of logarithms:

\(\log_5 250-\log_5 2=\log_5\dfrac{250}{2}=\log_5 125=3\).

Also, \(\log_4 2=\dfrac12\), because \(4^{1/2}=2\).

So the whole expression equals:

\(3+\dfrac12=\dfrac72\).

\(\log_4 2=\dfrac12\), because \(4^{1/2}=2\).

\(\log_4 8=\dfrac32\), because \(4^{3/2}=8\).

\(\log_3 \left(\dfrac13\right)=-1\).

Therefore,

\(\dfrac12+\dfrac32-(-1)=2+1=3\).

Use the property: \(2\log_{20}2=\log_{20}2^2=\log_{20}4\).

Then

\(\log_{20}4+\log_{20}5=\log_{20}(4\cdot5)=\log_{20}20=1\).

By the change of base formula:

\(\dfrac{\log_7 125}{\log_7 5}=\log_5 125=3\).

Also, \(\log_2 8=3\).

So

\(3+3=6\).

First:

\(\log_3 18-\log_3 2=\log_3 \dfrac{18}{2}=\log_3 9=2\).

Next, \(\log_{25}5=\dfrac12\), because \(25^{1/2}=5\).

Therefore,

\(2-\dfrac12=\dfrac32\).