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,

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

Add the common logarithms:

\(\lg 5+\lg 200=\lg(5\cdot200)=\lg1000=3\).

Also, \(\log_2 \left(\dfrac14\right)=-2\), because \(2^{-2}=\dfrac14\).

Then

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

Rewrite the first term:

\(2\log_6 3=\log_6 3^2=\log_6 9\).

Then

\(\log_6 9+\log_6 4=\log_6(9\cdot4)=\log_6 36=2\).

By the change of base formula:

\(\dfrac{\log_5 16}{\log_5 4}=\log_4 16=2\).

Also, \(\log_3 81=4\).

Therefore,

\(2+4=6\).

First:

\(\log_5 100-\log_5 4=\log_5 \dfrac{100}{4}=\log_5 25=2\).

Next, \(\log_{49}7=\dfrac12\), because \(49^{1/2}=7\).

Therefore,

\(2+\dfrac12=\dfrac52\).

Add the logarithms with the same base:

\(\log_{12}3+\log_{12}4=\log_{12}(3\cdot4)=\log_{12}12=1\).

Also, \(\log_4 \left(\dfrac{1}{16}\right)=-2\), because \(4^{-2}=\dfrac{1}{16}\).

Then

\(1-(-2)=3\).

Rewrite the first term:

\(3\log_{24}2=\log_{24}2^3=\log_{24}8\).

Then

\(\log_{24}8+\log_{24}3=\log_{24}(8\cdot3)=\log_{24}24=1\).

By the change of base formula:

\(\dfrac{\log_2 1000}{\log_2 10}=\log_{10}1000=3\).

Also, \(\log_3 9=2\).

Therefore,

\(3+2=5\).

First:

\(\log_5 40-\log_5 8=\log_5 \dfrac{40}{8}=\log_5 5=1\).

Also, \(\log_{100}10=\dfrac12\), because \(100^{1/2}=10\).

Therefore,

\(1-\dfrac12=\dfrac12\).

First add the common logarithms:

\(\lg 4+\lg 25=\lg(4\cdot25)=\lg100=2\).

Also, \(\log_3 \left(\dfrac{1}{27}\right)=-3\), because \(3^{-3}=\dfrac{1}{27}\).

Therefore,

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

Rewrite:

\(2\log_{20}2=\log_{20}4\).

Then

\(\log_{20}4+\log_{20}100=\log_{20}(4\cdot100)=\log_{20}400=2\).

By the change of base formula:

\(\dfrac{\log_4 81}{\log_4 9}=\log_9 81=2\).

Also, \(\log_2 64=6\).

Then

\(2+6=8\).