(a) $$ \ln e^{5}-2 \ln e^{6}=\square $$ (b) $$ \log _{14} 7+\log _{14} 2=\square $$

Question

UpStudy AI · Accepted Answer

Let's solve each part step by step.

(a) We start with the expression $$ \ln e^{5} - 2 \ln e^{6} $$.

Using the property of logarithms that states $$ \ln e^{x} = x $$, we can simplify:

$$
\ln e^{5} = 5
$$

Next, we simplify $$ 2 \ln e^{6} $$:

$$
2 \ln e^{6} = 2 \cdot 6 = 12
$$

Now we can substitute these values back into the expression:

$$
\ln e^{5} - 2 \ln e^{6} = 5 - 12 = -7
$$

Thus, the answer for part (a) is:

$$
\boxed{-7}
$$

(b) Now we will solve $$ \log_{14} 7 + \log_{14} 2 $$.

Using the property of logarithms that states $$ \log_{a} b + \log_{a} c = \log_{a} (bc) $$, we can combine the two logarithms:

$$
\log_{14} 7 + \log_{14} 2 = \log_{14} (7 \cdot 2) = \log_{14} 14
$$

Since $$ \log_{a} a = 1 $$, we have:

$$
\log_{14} 14 = 1
$$

Thus, the answer for part (b) is:

$$
\boxed{1}
$$
(a) $$ \ln e^{5}-2 \ln e^{6}=-7 $$ (b) $$ \log _{14} 7+\log _{14} 2=1 $$

(a) \( \ln e^{5}-2 \ln e^{6}=\square \) (b) \( \log _{14} 7+\log _{14} 2=\square \)