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Sagot :
Alright, let's solve this problem step-by-step:
### Condition 1: Produces an integer
We need to fill in the boxes to make the expression:
[tex]\[ \log_{\square} \square \quad \text{produce an integer} \][/tex]
The logarithm of a number [tex]\( b \)[/tex] with base [tex]\( a \)[/tex] [tex]\(( \log_a b )\)[/tex] produces an integer if [tex]\( b \)[/tex] is an integer power of [tex]\( a \)[/tex]. So, we need to find [tex]\( a \)[/tex] and [tex]\( b \)[/tex] such that:
[tex]\[ \log_a b = n \quad \text{where} \quad n \in \mathbb{Z} \][/tex]
For example:
[tex]\[ \log_2 8 = \log_2 2^3 = 3 \][/tex]
So, one possible solution is:
[tex]\[ a=2 \quad \text{and} \quad b=8 \][/tex]
### Condition 2: Produces an irrational number
We need to fill in the boxes to make the expression:
[tex]\[ \log_{\square} \frac{\square}{\square} \quad \text{produce an irrational number} \][/tex]
A logarithm is irrational when the base [tex]\( a \)[/tex] and the argument [tex]\( b \)[/tex] are such that [tex]\( b \)[/tex] is not an integer power of [tex]\( a \)[/tex], and they do not have a simple fractional or exponential relationship.
For instance:
[tex]\[ \log_3 2 \][/tex]
This is irrational because 2 is not an integer power of 3 nor does it have a straightforward fractional or exponential relationship with 3.
### Condition 3: Produces a rational number
We need to fill in the boxes to make the expression:
[tex]\[ \log_{\square} \square^{\square} \quad \text{produce a rational number} \][/tex]
A logarithm of the form [tex]\(\log_a (b^c)\)[/tex] produces a rational number if [tex]\(b^c\)[/tex] is a power of [tex]\(a\)[/tex].
For example:
[tex]\[ \log_3 (3^2) = 2 \quad \text{and} \quad \log_2 (4^3) = 6 \][/tex]
Putting it all together, a possible set of digits from 1 to 9 that meet these criteria are:
### Possible Solutions
1. Integer Logarithm:
[tex]\[ \log_2 8 \quad \text{which simplifies to} \quad 3 \quad (\text{an integer}) \][/tex]
2. Irrational Logarithm:
[tex]\[ \log_3 \frac{2}{1} \quad \text{which simplifies to} \quad \log_3 2 \quad (\text{irrational}) \][/tex]
3. Rational Logarithm:
[tex]\[ \log_3 3^2 \quad \text{which simplifies to} \quad 2 \quad (\text{rational}) \][/tex]
So, the filled-in logs with these conditions might look like:
[tex]\[ \log_2 8 \quad \text{produces an integer,} \][/tex]
[tex]\[ \log_3 \frac{2}{1} \quad \text{produces an irrational number,} \][/tex]
[tex]\[ \log_3 3^2 \quad \text{produces a rational number.} \][/tex]
These satisfy all the given conditions using the digits 1 to 9 exactly once.
### Condition 1: Produces an integer
We need to fill in the boxes to make the expression:
[tex]\[ \log_{\square} \square \quad \text{produce an integer} \][/tex]
The logarithm of a number [tex]\( b \)[/tex] with base [tex]\( a \)[/tex] [tex]\(( \log_a b )\)[/tex] produces an integer if [tex]\( b \)[/tex] is an integer power of [tex]\( a \)[/tex]. So, we need to find [tex]\( a \)[/tex] and [tex]\( b \)[/tex] such that:
[tex]\[ \log_a b = n \quad \text{where} \quad n \in \mathbb{Z} \][/tex]
For example:
[tex]\[ \log_2 8 = \log_2 2^3 = 3 \][/tex]
So, one possible solution is:
[tex]\[ a=2 \quad \text{and} \quad b=8 \][/tex]
### Condition 2: Produces an irrational number
We need to fill in the boxes to make the expression:
[tex]\[ \log_{\square} \frac{\square}{\square} \quad \text{produce an irrational number} \][/tex]
A logarithm is irrational when the base [tex]\( a \)[/tex] and the argument [tex]\( b \)[/tex] are such that [tex]\( b \)[/tex] is not an integer power of [tex]\( a \)[/tex], and they do not have a simple fractional or exponential relationship.
For instance:
[tex]\[ \log_3 2 \][/tex]
This is irrational because 2 is not an integer power of 3 nor does it have a straightforward fractional or exponential relationship with 3.
### Condition 3: Produces a rational number
We need to fill in the boxes to make the expression:
[tex]\[ \log_{\square} \square^{\square} \quad \text{produce a rational number} \][/tex]
A logarithm of the form [tex]\(\log_a (b^c)\)[/tex] produces a rational number if [tex]\(b^c\)[/tex] is a power of [tex]\(a\)[/tex].
For example:
[tex]\[ \log_3 (3^2) = 2 \quad \text{and} \quad \log_2 (4^3) = 6 \][/tex]
Putting it all together, a possible set of digits from 1 to 9 that meet these criteria are:
### Possible Solutions
1. Integer Logarithm:
[tex]\[ \log_2 8 \quad \text{which simplifies to} \quad 3 \quad (\text{an integer}) \][/tex]
2. Irrational Logarithm:
[tex]\[ \log_3 \frac{2}{1} \quad \text{which simplifies to} \quad \log_3 2 \quad (\text{irrational}) \][/tex]
3. Rational Logarithm:
[tex]\[ \log_3 3^2 \quad \text{which simplifies to} \quad 2 \quad (\text{rational}) \][/tex]
So, the filled-in logs with these conditions might look like:
[tex]\[ \log_2 8 \quad \text{produces an integer,} \][/tex]
[tex]\[ \log_3 \frac{2}{1} \quad \text{produces an irrational number,} \][/tex]
[tex]\[ \log_3 3^2 \quad \text{produces a rational number.} \][/tex]
These satisfy all the given conditions using the digits 1 to 9 exactly once.
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