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Fill in the missing values to make the equations true.

(a) [tex]\(\log_9 10 - \log_9 7 = \log_9 \square\)[/tex]

(b) [tex]\(\log_7 3 + \log_7 \square = \log_7 33\)[/tex]

(c) [tex]\(2 \log_9 2 = \log_9 \square\)[/tex]


Sagot :

Sure, let's go through each equation step-by-step and identify the missing values:

(a) [tex]\(\log_9 10 - \log_9 7 = \log_9 ?\)[/tex]

We can use the properties of logarithms here. Specifically, we use the property that states:
[tex]\[ \log_b A - \log_b B = \log_b \left(\frac{A}{B}\right) \][/tex]
Putting the given values into this form:
[tex]\[ \log_9 10 - \log_9 7 = \log_9 \left(\frac{10}{7}\right) \][/tex]
Thus, the missing value is:
[tex]\[ \boxed{1.4285714285714286} \][/tex]

(b) [tex]\(\log_7 3 + \log_7 \square = \log_7 33\)[/tex]

Again, we use another property of logarithms:
[tex]\[ \log_b A + \log_b B = \log_b (A \cdot B) \][/tex]
Setting this up with the given values:
[tex]\[ \log_7 3 + \log_7 ? = \log_7 (3 \cdot ?) \][/tex]
Since we know:
[tex]\[ \log_7 3 + \log_7 ? = \log_7 33 \][/tex]
We can infer:
[tex]\[ 3 \cdot ? = 33 \quad \Rightarrow \quad ? = \frac{33}{3} = 11 \][/tex]
So, the missing value is:
[tex]\[ \boxed{11} \][/tex]

(c) [tex]\(2 \log_9 2 = \log_9 \square\)[/tex]

Here, we can use the property that states:
[tex]\[ k \log_b A = \log_b (A^k) \][/tex]
Using [tex]\(k=2\)[/tex] and [tex]\(A=2\)[/tex], we get:
[tex]\[ 2 \log_9 2 = \log_9 (2^2) \][/tex]
So:
[tex]\[ 2^2 = 4 \][/tex]
Thus, the missing value is:
[tex]\[ \boxed{4} \][/tex]

In summary, we've filled in the missing values as follows:

(a) [tex]\(\log_9 10 - \log_9 7 = \log_9 \left(\frac{10}{7}\right) \implies \boxed{1.4285714285714286}\)[/tex]

(b) [tex]\(\log_7 3 + \log_7 11 = \log_7 33 \implies \boxed{11}\)[/tex]

(c) [tex]\(2 \log_9 2 = \log_9 4 \implies \boxed{4}\)[/tex]
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