Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
To simplify [tex]\(\ln (e x)\)[/tex], we can use the properties of logarithms. One essential property is:
[tex]\[ \ln(a \cdot b) = \ln(a) + \ln(b) \][/tex]
So, in the given expression [tex]\(\ln (e x)\)[/tex], we can decompose it as follows:
1. Identify the components inside the logarithm. Here [tex]\(a = e\)[/tex] and [tex]\(b = x\)[/tex].
2. Apply the logarithm product rule:
[tex]\[ \ln (e x) = \ln(e) + \ln(x) \][/tex]
3. Recognize that [tex]\(\ln(e) = 1\)[/tex] because [tex]\(e\)[/tex] is the base of the natural logarithm:
[tex]\[ \ln(e) = 1 \][/tex]
4. Substitute [tex]\(\ln(e)\)[/tex] with 1 in the equation:
[tex]\[ \ln(e x) = 1 + \ln(x) \][/tex]
Therefore, the expression [tex]\(\ln (e x)\)[/tex] simplifies to:
[tex]\[ 1 + \ln(x) \][/tex]
[tex]\[ \ln(a \cdot b) = \ln(a) + \ln(b) \][/tex]
So, in the given expression [tex]\(\ln (e x)\)[/tex], we can decompose it as follows:
1. Identify the components inside the logarithm. Here [tex]\(a = e\)[/tex] and [tex]\(b = x\)[/tex].
2. Apply the logarithm product rule:
[tex]\[ \ln (e x) = \ln(e) + \ln(x) \][/tex]
3. Recognize that [tex]\(\ln(e) = 1\)[/tex] because [tex]\(e\)[/tex] is the base of the natural logarithm:
[tex]\[ \ln(e) = 1 \][/tex]
4. Substitute [tex]\(\ln(e)\)[/tex] with 1 in the equation:
[tex]\[ \ln(e x) = 1 + \ln(x) \][/tex]
Therefore, the expression [tex]\(\ln (e x)\)[/tex] simplifies to:
[tex]\[ 1 + \ln(x) \][/tex]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.