(a) $$ 2 \ln e^{7}+\ln e^{4}=\square $$ (b) $$ \log _{6} 3-\log _{6} 18=\square $$

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(a) To solve the expression $$ 2 \ln e^{7}+\ln e^{4} $$, we can use the properties of logarithms. Specifically, the property that states $$ \ln e^x = x $$ for any real number $$ x $$.

So, we have:
$$ 2 \ln e^{7}+\ln e^{4} = 2 \cdot 7 + 4 $$
$$ = 14 + 4 $$
$$ = 18 $$

Therefore, the answer is $$ 18 $$.

(b) To solve the expression $$ \log _{6} 3-\log _{6} 18 $$, we can use the property of logarithms that states $$ \log_b a - \log_b c = \log_b \left(\frac{a}{c}\right) $$.

So, we have:
$$ \log _{6} 3-\log _{6} 18 = \log _{6} \left(\frac{3}{18}\right) $$
$$ = \log _{6} \left(\frac{1}{6}\right) $$

Since $$ \frac{1}{6} $$ is the same as $$ 6^{-1} $$, we can rewrite the logarithm as:
$$ \log _{6} \left(\frac{1}{6}\right) = \log _{6} 6^{-1} $$

Using the property $$ \log_b b^x = x $$, we get:
$$ \log _{6} 6^{-1} = -1 $$

Therefore, the answer is $$ -1 $$.
(a) $$ 2 \ln e^{7}+\ln e^{4}=18 $$ (b) $$ \log _{6} 3-\log _{6} 18=-1 $$

(a) \( 2 \ln e^{7}+\ln e^{4}=\square \) (b) \( \log _{6} 3-\log _{6} 18=\square \)

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