Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Let's go through each part of the questions step by step.
### Part A: Solve for [tex]\( x \)[/tex] in [tex]\( 5^x = 20 \)[/tex]
To solve for [tex]\( x \)[/tex] in the equation [tex]\( 5^x = 20 \)[/tex], we can use logarithms. Taking the logarithm of both sides (commonly natural logarithm [tex]\( \ln \)[/tex] or common logarithm [tex]\( \log \)[/tex]):
[tex]\[ \log(5^x) = \log(20) \][/tex]
Using the property of logarithms that [tex]\( \log(a^b) = b \log(a) \)[/tex]:
[tex]\[ x \log(5) = \log(20) \][/tex]
Now, solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\log(20)}{\log(5)} \][/tex]
### Part B: Solve for [tex]\( x \)[/tex] given [tex]\( \log_6 10 = x \)[/tex]
This is a straightforward logarithmic equation given in terms of a logarithm with base 6. To rewrite it:
[tex]\[ x = \log_6 10 \][/tex]
This can be interpreted directly without additional steps.
### Part C: Identify the error in the statement [tex]\( \log_3 8 = x \rightarrow 3^8 = x \)[/tex]
Let's first rewrite the given logarithmic equation correctly:
[tex]\[ \log_3 8 = x \][/tex]
This means that:
[tex]\[ 3^x = 8 \][/tex]
The error in the statement is in the misinterpretation of the logarithmic form. The original statement incorrectly asserts that [tex]\( 3^8 = x \)[/tex], whereas it should be:
[tex]\[ 3^x = 8 \][/tex]
### Part D: Given [tex]\( \log_4 7 \)[/tex]
Rewrite this logarithm in a more familiar base, such as natural log or common log, if desired. This doesn't require further solving since it's just an expression with a given value:
[tex]\[ x = \log_4 7 \][/tex]
### Part E: Calculate [tex]\( \log_4 10 + \log_4 2 \)[/tex]
Using the properties of logarithms, specifically the product rule [tex]\( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \)[/tex]:
[tex]\[ \log_4 10 + \log_4 2 = \log_4 (10 \cdot 2) = \log_4 20 \][/tex]
Simplify this equation:
The expression [tex]\( \log_4 10 + \log_4 2 \)[/tex] simplifies correctly to [tex]\( \log_4 20 \)[/tex]. The statement [tex]\( \log_4 10 + \log_4 2 = \log_4 5 \)[/tex] is incorrect.
### Part F: Common Log vs. Natural Log
Common logarithms are base 10, and are denoted as [tex]\( \log \)[/tex] without a base, while natural logarithms are base [tex]\( e \)[/tex] and are denoted as [tex]\( \ln \)[/tex].
For instance:
[tex]\[ \log(100) \approx 2 \][/tex]
[tex]\[ \ln(e^2) = 2 \][/tex]
Both types of logs are useful in different contexts, with common logs being useful in general calculations and natural logs frequently appearing in calculus and natural phenomena applications.
### Part A: Solve for [tex]\( x \)[/tex] in [tex]\( 5^x = 20 \)[/tex]
To solve for [tex]\( x \)[/tex] in the equation [tex]\( 5^x = 20 \)[/tex], we can use logarithms. Taking the logarithm of both sides (commonly natural logarithm [tex]\( \ln \)[/tex] or common logarithm [tex]\( \log \)[/tex]):
[tex]\[ \log(5^x) = \log(20) \][/tex]
Using the property of logarithms that [tex]\( \log(a^b) = b \log(a) \)[/tex]:
[tex]\[ x \log(5) = \log(20) \][/tex]
Now, solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\log(20)}{\log(5)} \][/tex]
### Part B: Solve for [tex]\( x \)[/tex] given [tex]\( \log_6 10 = x \)[/tex]
This is a straightforward logarithmic equation given in terms of a logarithm with base 6. To rewrite it:
[tex]\[ x = \log_6 10 \][/tex]
This can be interpreted directly without additional steps.
### Part C: Identify the error in the statement [tex]\( \log_3 8 = x \rightarrow 3^8 = x \)[/tex]
Let's first rewrite the given logarithmic equation correctly:
[tex]\[ \log_3 8 = x \][/tex]
This means that:
[tex]\[ 3^x = 8 \][/tex]
The error in the statement is in the misinterpretation of the logarithmic form. The original statement incorrectly asserts that [tex]\( 3^8 = x \)[/tex], whereas it should be:
[tex]\[ 3^x = 8 \][/tex]
### Part D: Given [tex]\( \log_4 7 \)[/tex]
Rewrite this logarithm in a more familiar base, such as natural log or common log, if desired. This doesn't require further solving since it's just an expression with a given value:
[tex]\[ x = \log_4 7 \][/tex]
### Part E: Calculate [tex]\( \log_4 10 + \log_4 2 \)[/tex]
Using the properties of logarithms, specifically the product rule [tex]\( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \)[/tex]:
[tex]\[ \log_4 10 + \log_4 2 = \log_4 (10 \cdot 2) = \log_4 20 \][/tex]
Simplify this equation:
The expression [tex]\( \log_4 10 + \log_4 2 \)[/tex] simplifies correctly to [tex]\( \log_4 20 \)[/tex]. The statement [tex]\( \log_4 10 + \log_4 2 = \log_4 5 \)[/tex] is incorrect.
### Part F: Common Log vs. Natural Log
Common logarithms are base 10, and are denoted as [tex]\( \log \)[/tex] without a base, while natural logarithms are base [tex]\( e \)[/tex] and are denoted as [tex]\( \ln \)[/tex].
For instance:
[tex]\[ \log(100) \approx 2 \][/tex]
[tex]\[ \ln(e^2) = 2 \][/tex]
Both types of logs are useful in different contexts, with common logs being useful in general calculations and natural logs frequently appearing in calculus and natural phenomena applications.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
