rewrite the logarithmic expression as a single logarithm with the same base. assume all expressions exist…

Explanation:

Step1: Recall log addition rule

The logarithm addition rule states that (\log_b M + \log_b N=\log_b(MN)) for the same base (b), and this can be extended to the sum of multiple logarithms with the same base. Also, recall that (\log_b a^n = n\log_b a), but here we can first use the addition rule for the sum of logarithms.

First, let's consider the sum (\log_y e^6+\log_y e+\log_y e^5). Since all logarithms have the same base (y), we can apply the property of logarithms: (\log_b A+\log_b B+\log_b C=\log_b(ABC)).

Step2: Apply the addition rule

First, identify the arguments of each logarithm: for (\log_y e^6), the argument is (e^6); for (\log_y e), the argument is (e); for (\log_y e^5), the argument is (e^5).

Using the rule (\log_b M+\log_b N+\log_b P=\log_b(MNP)), we multiply the arguments together:

(e^6\times e\times e^5)

When multiplying exponents with the same base, we add the exponents: (a^m\times a^n=a^{m + n}). So, (e^6\times e^1\times e^5=e^{6 + 1+5}=e^{12})

So, (\log_y e^6+\log_y e+\log_y e^5=\log_y(e^6\times e\times e^5)=\log_y e^{12})

We can also use the power rule of logarithms (\log_b a^n=n\log_b a), but in this case, we already combined the logarithms into a single logarithm. Alternatively, we can use the power rule on each term first:

(\log_y e^6 = 6\log_y e), (\log_y e=1\times\log_y e), (\log_y e^5 = 5\log_y e)

Then, (6\log_y e+\log_y e+5\log_y e=(6 + 1+5)\log_y e=12\log_y e)

But we can also express (12\log_y e=\log_y e^{12}) (by the power rule (\log_b a^n=n\log_b a) reversed, i.e., (n\log_b a=\log_b a^n))

Answer:

(\log_{y} e^{12}) (or (12\log_{y} e), but the form as a single logarithm with the argument as a single power of (e) is (\log_{y} e^{12}))