Question

rewrite the logarithmic expression as a single logarithm with the same base. assume all expressions exist and are well - defined. \\(\log_{y}e^{6}+\log_{y}e + \log_{y}e^{5}\\)

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 \(\log_{y} e^{12}\))