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Sagot :
Certainly! Let's solve each part step-by-step.
### Part (a) [tex]$\log _8 5-\log _8 11=\log _8 \square$[/tex]
To solve this, we will use the property of logarithms that states:
[tex]\[ \log_b(a) - \log_b(b) = \log_b \left(\frac{a}{b}\right) \][/tex]
Here, [tex]\(a = 5\)[/tex] and [tex]\(b = 11\)[/tex], so:
[tex]\[ \log_8 5 - \log_8 11 = \log_8 \left(\frac{5}{11}\right) \][/tex]
Therefore, the missing value is:
[tex]\[ \square = \frac{5}{11} \][/tex]
So, [tex]\(\log_8 5 - \log_8 11 = \log_8 \left(\frac{5}{11}\right) \)[/tex].
### Part (b) [tex]$\log _7 \square+\log _7 11=\log _7 33$[/tex]
To solve this, we will use the property of logarithms that states:
[tex]\[ \log_b(a) + \log_b(b) = \log_b (a \cdot b) \][/tex]
Here, [tex]\(b = 11\)[/tex] and the product is [tex]\(33\)[/tex]. We need to find [tex]\(a\)[/tex]:
We know that:
[tex]\[ \log_7 (a) + \log_7 (11) = \log_7 (33) \][/tex]
So, the value of [tex]\(a\)[/tex] must be:
[tex]\[ a \cdot 11 = 33 \][/tex]
[tex]\[ a = \frac{33}{11} \][/tex]
[tex]\[ a = 3 \][/tex]
Therefore, the missing value is:
[tex]\[ \square = 3 \][/tex]
So, [tex]\(\log_7 3 + \log_7 11 = \log_7 33 \)[/tex].
### Part (c) [tex]$\log _7 \frac{1}{25} = -2 \log _7 \square$[/tex]
To solve this, we will use the property of logarithms that states:
[tex]\[ \log_b \left(\frac{1}{a^2}\right) = -2 \log_b (a) \][/tex]
Here, the left-hand side is:
[tex]\[ \log_7 \left(\frac{1}{25}\right) \][/tex]
We need to find the value of [tex]\(a\)[/tex] such that:
[tex]\[ \log_7 \left(\frac{1}{25}\right) = -2 \log_7 (a) \][/tex]
We know that:
[tex]\[ \frac{1}{25} = \frac{1}{5^2} = 5^{-2} \][/tex]
So, we have:
[tex]\[ \log_7 \left(5^{-2}\right) = -2 \log_7 (5) \][/tex]
Therefore, the missing value is:
[tex]\[ \square = 5 \][/tex]
So, [tex]\(\log_7 \frac{1}{25} = -2 \log_7 5 \)[/tex].
### Summary of the Solutions:
(a) [tex]\(\log _8 5-\log _8 11=\log _8 \left(\frac{5}{11}\right)\)[/tex]
(b) [tex]\(\log _7 3+\log _7 11=\log _7 33\)[/tex]
(c) [tex]\(\log _7 \frac{1}{25}=-2 \log _7 5\)[/tex]
### Part (a) [tex]$\log _8 5-\log _8 11=\log _8 \square$[/tex]
To solve this, we will use the property of logarithms that states:
[tex]\[ \log_b(a) - \log_b(b) = \log_b \left(\frac{a}{b}\right) \][/tex]
Here, [tex]\(a = 5\)[/tex] and [tex]\(b = 11\)[/tex], so:
[tex]\[ \log_8 5 - \log_8 11 = \log_8 \left(\frac{5}{11}\right) \][/tex]
Therefore, the missing value is:
[tex]\[ \square = \frac{5}{11} \][/tex]
So, [tex]\(\log_8 5 - \log_8 11 = \log_8 \left(\frac{5}{11}\right) \)[/tex].
### Part (b) [tex]$\log _7 \square+\log _7 11=\log _7 33$[/tex]
To solve this, we will use the property of logarithms that states:
[tex]\[ \log_b(a) + \log_b(b) = \log_b (a \cdot b) \][/tex]
Here, [tex]\(b = 11\)[/tex] and the product is [tex]\(33\)[/tex]. We need to find [tex]\(a\)[/tex]:
We know that:
[tex]\[ \log_7 (a) + \log_7 (11) = \log_7 (33) \][/tex]
So, the value of [tex]\(a\)[/tex] must be:
[tex]\[ a \cdot 11 = 33 \][/tex]
[tex]\[ a = \frac{33}{11} \][/tex]
[tex]\[ a = 3 \][/tex]
Therefore, the missing value is:
[tex]\[ \square = 3 \][/tex]
So, [tex]\(\log_7 3 + \log_7 11 = \log_7 33 \)[/tex].
### Part (c) [tex]$\log _7 \frac{1}{25} = -2 \log _7 \square$[/tex]
To solve this, we will use the property of logarithms that states:
[tex]\[ \log_b \left(\frac{1}{a^2}\right) = -2 \log_b (a) \][/tex]
Here, the left-hand side is:
[tex]\[ \log_7 \left(\frac{1}{25}\right) \][/tex]
We need to find the value of [tex]\(a\)[/tex] such that:
[tex]\[ \log_7 \left(\frac{1}{25}\right) = -2 \log_7 (a) \][/tex]
We know that:
[tex]\[ \frac{1}{25} = \frac{1}{5^2} = 5^{-2} \][/tex]
So, we have:
[tex]\[ \log_7 \left(5^{-2}\right) = -2 \log_7 (5) \][/tex]
Therefore, the missing value is:
[tex]\[ \square = 5 \][/tex]
So, [tex]\(\log_7 \frac{1}{25} = -2 \log_7 5 \)[/tex].
### Summary of the Solutions:
(a) [tex]\(\log _8 5-\log _8 11=\log _8 \left(\frac{5}{11}\right)\)[/tex]
(b) [tex]\(\log _7 3+\log _7 11=\log _7 33\)[/tex]
(c) [tex]\(\log _7 \frac{1}{25}=-2 \log _7 5\)[/tex]
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